Nittka’s invariance criterion and Hilbert space valued parabolic equations in $L_p$

Abstract

Nittka gave an efficient criterion on a form defined on $L_2(\Omega )$ which implies that the associated semigroup is $L_p$-invariant for some given $p \in (1,\infty )$. We extend this criterion to the Hilbert space valued $L_2(\Omega ,H)$. As an application, we consider elliptic systems of purely second order. Our main result shows that the induced semigroup is $L_p$-contractive for all $p \in [p_-,p_+]$ for some $1< p_-< 2< p_+ < \infty$.

1 Introduction

Let $(S_t)_{t > 0}$ be a $C_0$-semigroup on $L_2(\Omega )$, with $\Omega \subset \mathbb {R}^d$, which is associated with a closed sesquilinear form. Ouhabaz [16] gave a convenient criterion on the form characterising when the semigroup is $L_\infty$-contractive. It is more complicated to describe $L_p$-contractivity if $1< p < \infty$. The reason is the fact that there is no explicit formula which describes the orthogonal projection from $L_2(\Omega )$ to the closed convex set $\{ u \in L_2(\Omega ): \Vert u\Vert _p \le 1 \}$ if $1< p < \infty$ and $p \ne 2$.

Nonetheless, Nittka [15] succeeded to overcome the difficulty by a structural analysis and developed an efficient criterion for the $L_p$-contractivity of a $C_0$-semigroup on $L_2(\Omega )$ which is associated with a form.

The purpose of this paper is twofold. Our first aim is to present Nittka’s result. We do this within the more general setting of the vector-valued space $L_2(\Omega ,H)$, where H is a Hilbert space. The result is the following.

Theorem 1.1

Let $(\Omega ,{{\mathcal {B}}},\mu )$ be a $\sigma$-finite measure space. Let H be a Hilbert space. Fix $p \in (1,\infty )$. Define

$$\begin{aligned} C = \{ u \in L_2(\Omega ,H) \cap L_p(\Omega ,H): \Vert u\Vert _{L_p(\Omega ,H)} \le 1 \}. \end{aligned}$$

Let P be the orthogonal projection of $L_2(\Omega ,H)$ onto C. Let ${{\mathcal {V}}}$ be a Hilbert space which is continuously and densely embedded in $L_2(\Omega ,H)$. Let $\mathfrak {a}:{{\mathcal {V}}}\times {{\mathcal {V}}}\rightarrow \mathbb {C}$ be a continuous elliptic sesquilinear form, let A be the operator in $L_2(\Omega ,H)$ associated with $\mathfrak {a}$, and let S be the semigroup generated by $-A$. Then the following are equivalent.

• (i) $\Vert S_t u\Vert _{L_p(\Omega ,H)} \le \Vert u\Vert _{L_p(\Omega ,H)}$ for all $u \in L_2(\Omega ,H) \cap L_p(\Omega ,H)$ and $t > 0$.

• (ii) $P {{\mathcal {V}}}\subset {{\mathcal {V}}}$ and ${\textrm{Re}}\,\mathfrak {a}(u, \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u) \ge 0$ for all $u \in {{\mathcal {V}}}$ with $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in {{\mathcal {V}}}$.

Our proof in Sect. 2 is slightly different from Nittka’s since we exploit the strict convexity of the space $L_p(\Omega ,H)$ for all $1< p < \infty$, which we prove in Appendix B.

Our second aim is to apply the criterion to purely second order Hilbert space valued elliptic operators with Neumann boundary conditions. They generate a contractive $C_0$-semigroup $(S_t)_{t > 0}$ on $L_2(\Omega ,H)$. Of particular interest are systems, that is, $H = \mathbb {C}^d$. In Sect. 3, we show that there is an interval $[p_-,p_+]$, with $1< p_-< 2< p_+ < \infty$, such that the semigroup S extends to a contractive $C_0$-semigroup on $L_p(\Omega ,H)$ for all $p \in [p_-,p_+]$. To prove the needed estimates, we use a chain rule formula, which is quite delicate and will be proved in Appendix A. In [2], related results are obtained and further interesting references are given. In the scalar case, our results may be compared with Cialdea–Maz’ya [7], who introduced an algebraic version of $L_p$-dissipativity and presented an algebraic characterisation for scalar-valued elliptic operators. This algebraic characterisation was refined by Carbonaro–Dragičević [6] and they used the result of Nittka to describe contractive $C_0$-semigroups on $L_p(\Omega )$ via the notion that they called p-ellipticity.

2 Nittka’s criterion for $L_p$-contractivity

Let $(\Omega ,{{\mathcal {B}}},\mu )$ be a $\sigma$-finite measure space. Let H be a Hilbert space. For all $p \in [1,\infty )$, we write $L_p = L_p(\Omega ,H)$. If $u \in L_p$, then we write $\Vert u\Vert _p = \Vert u\Vert _{L_p(\Omega ,H)}$ and $\Vert u\Vert _H :\Omega \rightarrow \mathbb {R}$ is the function from $\Omega$ into $\mathbb {R}$ such that

$$\begin{aligned} \Vert u\Vert _H(x) = \Vert u(x)\Vert _H \end{aligned}$$

for all $x \in \Omega$. Further we write ${{\mathcal {H}}}= L_2 = L_2(\Omega ,H)$. Throughout this paper, we fix $p \in (1,\infty )$. Define

$$\begin{aligned} C = \{ u \in {{\mathcal {H}}}\cap L_p: \Vert u\Vert _p \le 1 \}. \end{aligned}$$

Clearly C is convex and it follows from Fatou’s lemma that C is closed in ${{\mathcal {H}}}$. Let $P :{{\mathcal {H}}}\rightarrow C$ be the orthogonal projection. For all $u \in {{\mathcal {H}}}$, define

$$\begin{aligned} N(u) = \{ h \in {{\mathcal {H}}}: {\textrm{Re}}\,(h, v-u)_{{\mathcal {H}}}\le 0 \text{ for } \text{ all } v \in C \}. \end{aligned}$$

Then N(u) is a closed cone in ${{\mathcal {H}}}$ with $0 \in N(u)$. We state an easy property regarding C and N(u).

Lemma 2.1

For all $f \in {{\mathcal {H}}}$, there exist unique $u \in C$ and $h \in N(u)$ such that $f = u + h$. Actually, $u = Pf$.

Proof

Let $u = Pf$ and $h = f - Pf$. Then $u \in C$ and ${\textrm{Re}}\,(h, v-u)_{{\mathcal {H}}}= {\textrm{Re}}\,(f - Pf, v - Pf)_{{\mathcal {H}}}\le 0$ for all $v \in C$. This proves existence. Let also ${\tilde{u}} \in C$ and ${\tilde{h}} \in N({\tilde{u}})$ be such that $f = {\tilde{u}} + {\tilde{h}}$. Then ${\textrm{Re}}\,(h, {\tilde{u}} - u)_{{\mathcal {H}}}\le 0$ and similarly ${\textrm{Re}}\,({\tilde{h}}, u - \tilde{u})_{{\mathcal {H}}}\le 0$. Hence ${\textrm{Re}}\,(h - {\tilde{h}}, {\tilde{u}} - u)_{{\mathcal {H}}}\le 0$. Since $h - {\tilde{h}} = {\tilde{u}} - u$, this implies that ${\textrm{Re}}\,\Vert h - {\tilde{h}}\Vert _{{\mathcal {H}}}^2 \le 0$ and the statement follows. $\square$

If $u :\Omega \rightarrow H$ is a function, then define ${\textrm{sgn}}\,u :\Omega \rightarrow H$ by

$$\begin{aligned} ({\textrm{sgn}}\,u)(x) = \left\{ \begin{array}{ll} \frac{1}{\Vert u(x)\Vert _H} \, u(x) &{} \text{ if } u(x) \ne 0, \\ 0 &{} \text{ if } u(x) = 0. \end{array} \right. \end{aligned}$$

Note that ${\textrm{sgn}}\,u = \lim _{\varepsilon \downarrow 0} u_\varepsilon$ pointwise, where the function $u_\varepsilon :\Omega \rightarrow H$ is defined by $u_\varepsilon (x) = \frac{1}{\sqrt{\Vert u(x)\Vert _H^2 + \varepsilon }} \, u(x)$. Hence ${\textrm{sgn}}\,u$ is (Bochner) measurable if u is (Bochner) measurable. Then also $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u$ is (Bochner) measurable whenever u is (Bochner) measurable.

For a description of N(u), we use that $L_p(\Omega ,H)$ is strictly convex. See Appendix B for a proof. Recall that a Banach space E is called strictly convex if for all $\xi ,\eta \in E$ with $\Vert \xi \Vert _E = 1 = \Vert \eta \Vert _E$ and $\xi \ne \eta$, it follows that $\Vert \xi + \eta \Vert _E < 2$.

Lemma 2.2

Let E be a Banach space. Assume that $E^*$ is strictly convex and let $x \in E$. Then there exists a unique $f \in E^*$ such that ${\textrm{Re}}\,f(x) = \Vert x\Vert _E^2 = \Vert f\Vert _{E^*}^2$. This unique f satisfies $f(x) = \Vert x\Vert _E^2$.

Proof

The existence of an $f \in E^*$ such that $f(x) = \Vert x\Vert _E^2 = \Vert f\Vert _{E^*}^2$ is a well-known consequence of the Hahn–Banach theorem. For the uniqueness, we may assume without loss of generality that $\Vert x\Vert _E = 1$. Suppose also $g \in E^*$ with ${\textrm{Re}}\,g(x) = \Vert x\Vert _E^2 = \Vert g\Vert _{E^*}^2$ and $g \ne f$. Define $h = \frac{1}{2} \, (f + g)$. Then $\Vert h\Vert _{E^*} < 1$ by the strict convexity of $E^*$. But then

$$\begin{aligned} 1 = {\textstyle \frac{1}{2}} \, {\textrm{Re}}\,(f(x) + g(x)) = {\textrm{Re}}\,h(x) \le \Vert h\Vert _{E^*} \, \Vert x\Vert _E < 1. \end{aligned}$$

This is a contradiction. $\square$

Now we are able to give a characterisation for N(u).

Proposition 2.3

Let $u \in C$. Then the following are equivalent.

• (i) $N(u) \ne \{ 0 \}$.

• (ii) $\Vert u\Vert _p = 1$ and $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in {{\mathcal {H}}}$.

If these conditions are valid, then $N(u) = \{ t \, \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u: t \in [0,\infty ) \}$.

Proof

‘(ii)$\Rightarrow$(i)’. Let $v \in C$. Then the Cauchy–Schwarz inequality and the Hölder inequality give

$$\begin{aligned} {\textrm{Re}}\,(\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u, v)_{{\mathcal {H}}}\le & {} \int \limits _\Omega \Vert u\Vert _H^{p-1} \, \Vert v\Vert _H \\\le & {} \Big ( \int \limits _\Omega \Vert u\Vert _H^{(p-1) p'} \Big )^{1/p'} \, \Vert v\Vert _p \\\le & {} \Vert u\Vert _p^{p/p'} = 1 = {\textrm{Re}}\,(\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u, u)_{{\mathcal {H}}}. \end{aligned}$$

So $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in N(u)$ and $\{ t \, \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u: t \in [0,\infty ) \} \subset N(u)$. In particular, $N(u) \ne \{ 0 \}$.

‘(i)$\Rightarrow$(ii)’. We first show that $\Vert u\Vert _p = 1$. Suppose that $\Vert u\Vert _p < 1$. Let $h \in N(u)$. Let $w \in L_2 \cap L_p$ with $\Vert w\Vert _p \le 1 - \Vert u\Vert _p$. Then $u + w \in C$. Since $h \in N(u)$, one deduces that ${\textrm{Re}}\,(h,w)_{{\mathcal {H}}}\le 0$. Because $L_2 \cap L_p$ is dense in ${{\mathcal {H}}}$, it follows that $h = 0$. Hence (i) implies that $\Vert u\Vert _p = 1$.

Next let $h \in N(u)$ with $h \ne 0$. If $v \in L_2(\Omega ,H) \cap L_p(\Omega ,H)$, then ${\textrm{Re}}\,(h,v)_{{\mathcal {H}}}\le \Vert v\Vert _p \, {\textrm{Re}}\,(h,u)_{{\mathcal {H}}}$. So $h \in L_{p'}(\Omega ,H)$ and $\Vert h\Vert _{p'} \le {\textrm{Re}}\,(h,u)_{{\mathcal {H}}}$. If ${\textrm{Re}}\,(h,u)_{{\mathcal {H}}}= 0$, then $\Vert h\Vert _{p'} = 0$ and $h = 0$, which is a contradiction. So ${\textrm{Re}}\,(h,u)_{{\mathcal {H}}}\ne 0$. Multiplying h with a strictly positive constant, we may assume that ${\textrm{Re}}\,(h,u)_{{\mathcal {H}}}= 1$. Then

$$\begin{aligned} \Vert h\Vert _{p'} \le 1 = {\textrm{Re}}\,(h,u)_{{\mathcal {H}}}\le \Vert h\Vert _{p'} \, \Vert u\Vert _p \le \Vert h\Vert _{p'}. \end{aligned}$$

So $\Vert h\Vert _{p'} = 1 = {\textrm{Re}}\,(h,u)_{{\mathcal {H}}}= {\textrm{Re}}\,\langle h,u \rangle _{L_{p'} \times L_p}$. We proved that

$$\begin{aligned} {\textrm{Re}}\,\langle h,u \rangle _{L_{p'} \times L_p} = \Vert h\Vert _{p'}^2 = \Vert u\Vert _p^2. \end{aligned}$$

On the other hand,

$$\begin{aligned} \int \limits _\Omega \Big \Vert \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \Big \Vert _H^{p'} = \int \limits _\Omega \Vert u\Vert _H^{(p-1) p'} = \int \limits _\Omega \Vert u\Vert _H^p = 1, \end{aligned}$$

so $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in L_{p'}$ and furthermore

$$\begin{aligned} {\textrm{Re}}\,\langle \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u,u \rangle _{L_{p'} \times L_p} = \int \limits _\Omega \Vert u\Vert _H^p = 1 = \Big \Vert \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \Big \Vert _{p'}^2 = \Vert u\Vert _p^2. \end{aligned}$$

By Lemma 2.2, we obtain that $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u = h \in {{\mathcal {H}}}$. Moreover, $N(u) \subset \{ t \, \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u: t \in [0,\infty ) \}$. $\square$

Let ${{\mathcal {V}}}$ be a Hilbert space that is continuously and densely embedded in ${{\mathcal {H}}}= L_2(\Omega ,H)$. Let $\mathfrak {a}:{{\mathcal {V}}}\times {{\mathcal {V}}}\rightarrow \mathbb {C}$ be a sesquilinear form such that $\mathfrak {a}$ is continuous, that is, there is an $M > 0$ such that

$$\begin{aligned} |\mathfrak {a}(u,v)| \le M \, \Vert u\Vert _{{\mathcal {V}}}\, \Vert v\Vert _{{\mathcal {V}}}\end{aligned}$$

for all $u,v \in {{\mathcal {V}}}$, and $\mathfrak {a}$ is elliptic, that is, there are $\mu > 0$ and $\omega \in \mathbb {R}$ such that

$$\begin{aligned} {\textrm{Re}}\,\mathfrak {a}(u,u) + \omega \, ||u\Vert _{{\mathcal {H}}}^2 \ge \mu \, \Vert u\Vert _{{\mathcal {V}}}^2 \end{aligned}$$

(1)

for all $u \in {{\mathcal {V}}}$. Then there is a unique operator A in ${{\mathcal {H}}}$ whose graph is

$$\begin{aligned} {\textrm{graph}}\,(A) = \{ (u,f): u \in {{\mathcal {V}}}, \; f \in {{\mathcal {H}}}, \text{ and } \mathfrak {a}(u,v) = (f,v)_{{\mathcal {H}}} \text{ for } \text{ all } v \in {{\mathcal {V}}}\}. \end{aligned}$$

We call A the operator associated with the form $\mathfrak {a}$. Then $-A$ generates a holomorphic $C_0$-semigroup $(S_t)_{t > 0}$ in ${{\mathcal {H}}}$ satisfying $\Vert S_t\Vert _{{{\mathcal {H}}}\rightarrow {{\mathcal {H}}}} \le e^{\omega t}$ for all $t > 0$, where $\omega$ is as in (1).

Proof of Theorem 1.1

‘(i)$\Rightarrow$(ii)’. It follows from [16, Theorem 2.2 1)$\Rightarrow$2)], that $P {{\mathcal {V}}}\subset {{\mathcal {V}}}$ and ${\textrm{Re}}\,\mathfrak {a}(Pf, f - Pf) \ge 0$ for all $f \in {{\mathcal {V}}}$. Since $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in L_2(\Omega ,H)$ by assumption, one deduces that

$$\begin{aligned} \int \limits _\Omega \Vert u\Vert _H^p = \int \limits _\Omega ( \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u, \Vert u\Vert _H \, {\textrm{sgn}}\,u)_H < \infty . \end{aligned}$$

So $u \in L_p$. Without loss of generality, we may assume that $\Vert u\Vert _p = 1$. Then $u \in C$ and $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in {{\mathcal {H}}}$, so $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in N(u)$ by Proposition 2.3. Set $f = u + \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in {{\mathcal {V}}}\subset {{\mathcal {H}}}$. Then the uniqueness of Lemma 2.1 gives $u = Pf$. Consequently ${\textrm{Re}}\,\mathfrak {a}(u, \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u) = {\textrm{Re}}\,\mathfrak {a}(Pf, f - Pf) \ge 0$.

‘(ii)$\Rightarrow$(i)’. Let $v \in {{\mathcal {V}}}$. We shall show that ${\textrm{Re}}\,\mathfrak {a}(Pv, v - Pv) \ge 0$. If $v \in C$, then this is trivial, so we may assume that $v \not \in C$. Set $u = Pv$ and $h = v - u$. Then $h \in N(u)$ by Lemma 2.1. Also $h \ne 0$, so Proposition 2.3 implies that $\Vert u\Vert _p = 1$ and $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u \in {{\mathcal {H}}}$. Moreover, there exists a $t \in [0,\infty )$ such that $h = t \, \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u$. Then $t \ne 0$. Since $u = Pv \in P {{\mathcal {V}}}\subset {{\mathcal {V}}}$ by assumption, one deduces that $\Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u = \frac{1}{t} (v-u) \in {{\mathcal {V}}}$. Hence ${\textrm{Re}}\,\mathfrak {a}(Pv, v - Pv) = t {\textrm{Re}}\,\mathfrak {a}(u, \Vert u\Vert _H^{p-1} \, {\textrm{sgn}}\,u) \ge 0$. Now it follows from [17, Theorem 2.2 2)$\Rightarrow$1)], that $S_t C \subset C$ for all $t > 0$. This obviously implies statement (i). $\square$

3 Application to Hilbert space valued parabolic problems

Let $\Omega \subset \mathbb {R}^d$ be an open set. Let H be a separable Hilbert space. For all $k,l \in \{ 1,\ldots ,d \}$, let $c_{kl} :\Omega \rightarrow {{\mathcal {L}}}(H)$ be a bounded function such that $x \mapsto (c_{kl}(x) \, \xi , \eta )_H$ is measurable from $\Omega$ into $\mathbb {C}$ for all $\xi ,\eta \in H$. Let $\mu > 0$. We assume that

$$\begin{aligned} {\textrm{Re}}\,\sum _{k,l=1}^d (c_{kl}(x) \, \xi _l, \xi _k)_H \ge \mu \sum _{k=1}^d \Vert \xi _k\Vert _H^2 \end{aligned}$$

for all $x \in \Omega$ and $\xi _1,\ldots ,\xi _d \in H$. Further let $M > 0$ be such that

$$\begin{aligned} \sum _{k=1}^d \Big \Vert \sum _{l=1}^d c_{kl}(x) \, \xi _l \Big \Vert _H^2 \le M^2 \, \sum _{k=1}^d \Vert \xi _k\Vert _H^2 \end{aligned}$$

for all $x \in \Omega$ and $\xi _1,\ldots ,\xi _d \in H$. For simplicity, we consider Neumann boundary conditions. Define ${{\mathcal {V}}}= H^1(\Omega ,H)$ and $\mathfrak {a}:{{\mathcal {V}}}\times {{\mathcal {V}}}\rightarrow \mathbb {C}$ by

$$\begin{aligned} \mathfrak {a}(u,v) = \sum _{k,l=1}^d \int \limits _\Omega (c_{kl}(x) \, (\partial _l u)(x), (\partial _k v)(x) )_H \, dx. \end{aligned}$$

Then $\mathfrak {a}$ is a continuous elliptic sesquilinear form. Let A be the operator associated with $\mathfrak {a}$ and let S be the semigroup generated by $-A$.

Theorem 3.1

Let $p \in (1,\infty )$ and suppose that

$$\begin{aligned} \frac{\mu }{M} \ge 2 \Big | \frac{p-2}{p} \Big | + \Big | \frac{p-2}{p} \Big |^2. \end{aligned}$$

Then S extends consistently to a contraction semigroup in $L_p(\Omega ,H)$.

Note that the condition in Theorem 3.1 is invariant by taking the dual exponent, that is, if $p \in (1,\infty )$ satisfies the condition, then so does q, where $\frac{1}{p} + \frac{1}{q} = 1$. If one is satisfied with some small interval, $1< p_-< 2< p_+ < \infty$ such that S is $L_p$-contractive for all $p \in [p_-,p_+]$, then a significantly easier and less technical proof than the following can be given.

Proof

Using duality, without loss of generality, we may assume that $p > 2$. We argue as in [7, 12], and [11]. Let $u \in H^1(\Omega ,H)$. For all $n \in \mathbb {N}$, define

$$\begin{aligned} v_n = (\Vert u\Vert _H^{\frac{p-2}{2}} \wedge n) \, u, \quad w_n = (\Vert u\Vert _H^{p-2} \wedge n^2) \, u, \quad \text{ and } \quad \chi _n = \mathbbm {1}_{[\Vert u\Vert _H^{p-2} < n^2]}. \end{aligned}$$

If follows from Proposition A.1 (cf. [4, Theorems 3.3 and 4.2]) that $v_n,w_n \in H^1(\Omega ,H)$ with

$$\begin{aligned} \nabla v_n= & {} n \, (\mathbbm {1}- \chi _n) \, \nabla u + \chi _n \, \Vert u\Vert _H^{\frac{p-2}{2}} \Big ( \nabla u + {\textstyle \frac{p-2}{2}} \, (\nabla \Vert u\Vert _H) \, {\textrm{sgn}}\,u \Big ), \text{ and } \nonumber \\ \nabla w_n= & {} n^2 \, (\mathbbm {1}- \chi _n) \, \nabla u + \chi _n \, \Vert u\Vert _H^{p-2} \Big ( \nabla u + (p-2) \, (\nabla \Vert u\Vert _H) \, {\textrm{sgn}}\,u \Big ) . \end{aligned}$$

(2)

Note that $\chi _n \, \Vert v_n\Vert _H = \chi _n \, \Vert u\Vert _H^{p/2}$. Hence

$$\begin{aligned} (\mathbbm {1}- \chi _n) \, \nabla u= & {} \frac{1}{n} \, (\mathbbm {1}- \chi _n) \, \nabla v_n, \\ \chi _n \, \Vert v_n\Vert _H^{\frac{p-2}{p}} \, \nabla u= & {} \chi _n \, \Big ( \nabla v_n - {\textstyle \frac{p-2}{p}} \, (\nabla \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n \Big ), \text{ and } \\ \nabla w_n= & {} n \, (\mathbbm {1}- \chi _n) \, \nabla v_n + \chi _n \, \, \Vert v_n\Vert _H^{\frac{p-2}{p}} \Big ( \nabla v_n + {\textstyle \frac{p-2}{p}} \, (\nabla \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n \Big ). \end{aligned}$$

Therefore, for almost all $x \in \Omega$, one obtains

$$\begin{aligned}{} & {} { \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \partial _l u, \partial _k w_n )_H } \\{} & {} \quad = (\mathbbm {1}- \chi _n) \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \partial _l u, n \, \partial _k v_n )_H\\{} & {} \qquad + \chi _n \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \Vert v_n\Vert _H^{\frac{p-2}{p}} \, \partial _l u, \Big ( \partial _k v_n + {\textstyle \frac{p-2}{p}} \, (\partial _k \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n \Big ) )_H \\{} & {} \quad = (\mathbbm {1}- \chi _n) \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \partial _l v_n, \partial _k v_n )_H\\{} & {} \qquad + \chi _n \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \Big ( \partial _l v_n - {\textstyle \frac{p-2}{p}} \, (\partial _l \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n \Big ),\\{} & {} \quad \quad \Big ( \partial _k v_n + {\textstyle \frac{p-2}{p}} \, (\partial _k \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n \Big ) )_H \\{} & {} \quad = \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \partial _l v_n, \partial _k v_n )_H + {\textstyle \frac{p-2}{p}} \, \chi _n \, {\textrm{Re}}\,( (c_{kl} - c_{lk}^*) \, \partial _l v_n, (\partial _k \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n )_H\\{} & {} \qquad - \Big ( {\textstyle \frac{p-2}{p}} \Big )^2 \, {\textrm{Re}}\,(c_{kl} \, (\partial _l \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n, (\partial _k \Vert v_n\Vert _H) \, {\textrm{sgn}}\,v_n )_H \\{} & {} \quad \ge \Big ( \mu - 2 M \, {\textstyle \frac{p-2}{p}} - M \, \Big ( {\textstyle \frac{p-2}{p}} \Big )^2 \Big ) \sum _{k=1}^d \Vert \partial _k v_n\Vert _H^2, \end{aligned}$$

where we used that $\sum _{k=1}^d | \partial _k \Vert v_n\Vert _H |^2 \le \sum _{k=1}^d \Vert \partial _k v_n\Vert _H^2$ and $\Vert {\textrm{sgn}}\,v_n\Vert _H \le 1$ by Lemma A.5. By the assumption on p, we obtain that

$$\begin{aligned} \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \partial _l u, \partial _k w_n )_H \ge 0 \end{aligned}$$

almost everywhere. This is for all $n \in \mathbb {N}$.

Next take the limit $n \rightarrow \infty$ and use (2). Then

$$\begin{aligned} \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \partial _l u, \Vert u\Vert _H^{p-2} \, (\partial _k u + (p-2) \, (\partial _k \Vert u\Vert _H) \, {\textrm{sgn}}\,u ))_H \ge 0 \end{aligned}$$

(3)

almost everywhere.

We assume from now on that in addition $\Vert u\Vert _H^{p-2} \, u \in {{\mathcal {V}}}$. Let $k \in \{ 1,\ldots ,d \}$. We shall show that $\partial _k (\Vert u\Vert _H^{p-2} \, u) = f_k$ almost everywhere, where

$$\begin{aligned} f_k = \Vert u\Vert _H^{p-2} \, \Big ( \partial _k u + (p-2) \, (\partial _k \Vert u\Vert _H) \, {\textrm{sgn}}\,u \Big ). \end{aligned}$$

Write $r = 2 \, \frac{p-1}{p-2} \in (2,\infty )$ and let $q \in (1,2)$ be such that $\frac{1}{2} + \frac{1}{r} = \frac{1}{q}$. Since $\Vert u\Vert _H^{p-1} \in L_2(\Omega )$, one deduces that $\int _\Omega (\Vert u\Vert _H^{p-2})^r = \int _\Omega (\Vert u\Vert _H^{p-1})^2 < \infty$. So $\Vert u\Vert _H^{p-2} \in L_r(\Omega )$. If $n \in \mathbb {N}$, then $n^2 \, (\mathbbm {1}- \chi _n) \, \Vert \partial _k u\Vert _H \le (\mathbbm {1}- \chi _n) \, \Vert u\Vert _H^{p-2} \, \Vert \partial _k u\Vert _H$ and hence (2) gives

$$\begin{aligned} \Vert \partial _k w_n\Vert _H \le \Vert u\Vert _H^{p-2} \Big ( \Vert \partial _k u\Vert _H + (p-2) \, \Big | \partial _k \Vert u\Vert _H \Big | \Big ). \end{aligned}$$

(4)

Note that the right hand side of (4) does not depend on n and is an element of $L_q(\Omega )$. Also $\lim \partial _k w_n = f_k$ almost everywhere. Hence $\lim \partial _k w_n = f_k$ in $L_q(\Omega ,H)$. It is easy to see that $\lim w_n = \Vert u\Vert _H^{p-2} \, u$ in $L_2(\Omega ,H)$ since $\Vert u\Vert _H^{p-2} \, u \in L_2(\Omega ,H)$. Let $\varphi \in C_c^\infty (\Omega ,H)$. Then

$$\begin{aligned} \int \limits _\Omega (\partial _k (\Vert u\Vert _H^{p-2} \, u), \varphi )_H= & {} - \int \limits _\Omega (\Vert u\Vert _H^{p-2} \, u, \partial _k \varphi )_H = - \lim _{n \rightarrow \infty } \int \limits _\Omega (w_n, \partial _k \varphi )_H\\= & {} \lim _{n \rightarrow \infty } \int \limits _\Omega (\partial _k w_n, \varphi )_H = \int \limits _\Omega (f_k, \varphi )_H. \end{aligned}$$

So $\partial _k (\Vert u\Vert _H^{p-2} \, u) = f_k$ almost everywhere.

Finally (3) implies that

$$\begin{aligned} \sum _{k,l=1}^d {\textrm{Re}}\,(c_{kl} \, \partial _l u, \partial _k (\Vert u\Vert _H^{p-2} \, u))_H \ge 0 \end{aligned}$$

almost everywhere. Integrating over $\Omega$ gives ${\textrm{Re}}\,\mathfrak {a}(u,\Vert u\Vert _H^{p-2} \, u) \ge 0$. Now apply Theorem 1.1.

By the way, with some more work, one can show that $\lim _{n \rightarrow \infty } \partial _k w_n = f_k$ in $L_2(\Omega ,H)$. $\square$

We comment on related results concerning the extension of S to a not necessarily contractive $C_0$-semigroup on $L_p$. Hofmann, Mayboroda, and McIntosh [14] showed for $H=\mathbb {C}$, $\Omega =\mathbb {R}^d$, and $d\ge 3$ that the semigroup S can be extended to a $C_0$-semigroup on $L_p(\Omega )$ if $p\in [\frac{2d}{d+2},\frac{2d}{d-2}]$. Conversely, for each $p\in (1,\frac{2d}{d+2})\cup (\frac{2d}{d-2},\infty )$, they construct an elliptic operator such that the associated semigroup $(S_t)_{t>0}$ cannot be extended consistently to a bounded semigroup $L_p(\mathbb {R}^d)$. The extension results are based on off-diagonal Davies–Gaffney estimates; cf. also [8, Theorem 25], and [5, Section 3.1].

Davies had already pointed out in the introduction of [9] that his proof of [8, Theorem 25] extends to the vector-valued case. Moreover, by [9, Theorem 10], for each $p\in (1,\frac{2d}{d+2})\cup (\frac{2d}{d-2},\infty )$, there exists an elliptic system with $H=\mathbb {C}^d$, $\Omega =\mathbb {R}^d$, $d\ge 3$, and with real symmetric coefficients such that the operator $S_t$ does not continuously extend to $L_p$ for any $t>0$.

We shall give a corresponding extension result for our setting, which we obtain readily from standard estimates and tracing Auscher’s proof of [5, Proposition 3.2].

Theorem 3.2

Suppose $\Omega =\mathbb {R}^d$ or $\Omega \subset \mathbb {R}^d$ is open and Lipschitz. Then the semigroup S extends to a $C_0$-semigroup with growth bound 0 on $L_p(\Omega ,H)$ for all $p\in (\frac{2d}{d+2},\frac{2d}{d-2})$ if $d\ge 3$ and for all $p\in (1,\infty )$ if $d\in \{1,2\}$.

Proof

We outline the arguments for $d\ge 3$. Let $\omega >0$. Since $\mathfrak {a}$ is elliptic for this choice of $\omega$, by [18, Lemma 3.6.2 (3.60)] there exists a $c>0$ such that $\Vert e^{-\omega t} \, S_t u\Vert _{{\mathcal {V}}}\le c \, t^{-1/2} \, \Vert u\Vert _2$ for all $u\in L_2(\Omega ,H)$ and $t > 0$. Combining this with the Sobolev embedding ${{\mathcal {V}}}\hookrightarrow L_{2d/(d-2)}$, we obtain that there exists a $C>0$ such that

$$\begin{aligned} \Vert e^{-\omega t} \, S_t u\Vert _{2d/(d-2)} \le C \, t^{-1/2} \, \Vert u\Vert _2 \end{aligned}$$

for all $u\in L_2(\Omega ,H)$ and $t > 0$. Then, by duality,

$$\begin{aligned} \Vert e^{-\omega t} \, S_t^* u\Vert _2 \le C \, t^{-1/2} \, \Vert u\Vert _{2d/(d+2)} \end{aligned}$$

for all $u \in L_2 \cap L_{2d/(d+2)}$ and $t > 0$.

Next, it follows from inspection of the proof of [5, Proposition 3.2] that the parts (2) and (3) of [5, Proposition 3.2] extend to the vector-valued case and general open sets $\Omega$, and are applicable to the $C_0$-semigroup $(T_t)_{t>0}$ given by $T_t = e^{-\omega t} \, S_t^*$. For the extension of part (2), one needs $L_2$–$L_2$ off-diagonal estimates that can be obtained, for example, as in [3, Theorem 4.2]. Moreover, the vector-valued version of the Riesz–Thorin theorem follows from [13, Lemma 2.6]. Let $q \in (\frac{2d}{d+2},2)$. By the extension of [5, Proposition 3.2 (2)], we obtain that T satisfies $L_q$–$L_2$ off-diagonal estimates, which implies by the extension of [5, Proposition 3.2 (3)] that T is uniformly bounded in $L_q$. Dualizing again, we obtain the statement for all $p \in (2,\frac{2d}{d-2})$. By considering the adjoint form, applying the result for $p>2$, and taking the dual, we obtain the statement for $p\in (\frac{2d}{d+2},2)$. $\square$

Remark 3.3

We comment on the admissible ranges for p in Theorems 3.1 and 3.2. Remarkably, it is possible that the range given in Theorem 3.1 for contractive extensions is larger than the one given in Theorem 3.2 for extensions with growth bound 0. For example, this occurs if $\frac{\mu }{M} \ge \frac{1}{2}$, say, and d is sufficiently large.

Appendices

A. The derivative of a truncation

Let $\Omega \subset \mathbb {R}^d$ be an open set. Let H be a Hilbert space. The principal aim in this section is to prove the following chain rule.

Proposition A.1

Let $\alpha > 0$ and $M > 0$. Let $u \in H^1(\Omega ,H)$. Define $v = (\Vert u\Vert _H^\alpha \wedge M) \, u$. Then $v \in H^1(\Omega ,H)$ and

$$\begin{aligned} \partial _k v = \alpha \, \mathbbm {1}_{[\Vert u\Vert _H^\alpha < M]} \, \Vert u\Vert _H^\alpha \, {\textrm{Re}}\,({\textrm{sgn}}\,u, \partial _k u)_H \, {\textrm{sgn}}\,u + (\Vert u\Vert _H^{\alpha } \wedge M) \, \partial _k u \end{aligned}$$

for all $k \in \{ 1,\ldots ,d \}$.

The proof involves some work. We use the following approximation by smooth functions.

Lemma A.2

The space $C^\infty (\Omega ,H) \cap H^1(\Omega ,H)$ is dense in $H^1(\Omega ,H)$.

Proof

This follows as in the scalar case in [1, Theorem 3.17]. $\square$

For an approximation argument, the next lemma is useful.

Lemma A.3

For all $n \in \mathbb {N}$, let $u_n \in H^1(\Omega ,H)$. Let $u,g_1,\ldots ,g_d \in L_2(\Omega ,H)$. Suppose that $\lim u_n = u$ in $L_2(\Omega ,H)$ and $\lim \partial _k u_n = g_k$ in $L_2(\Omega ,H)$ for all $k \in \{ 1,\ldots ,d \}$. Then $u \in H^1(\Omega ,H)$ and $\partial _k u = g_k$ for all $k \in \{ 1,\ldots ,d \}$.

Proof

Let $\varphi \in C_c^\infty (\Omega )$. Let $k \in \{ 1,\ldots ,d \}$. Then $- \int _\Omega u_n \, \partial _k \varphi = \int _\Omega (\partial _k u_n) \, \varphi$ for all $n \in \mathbb {N}$. Then the lemma follows by taking the limit $n \rightarrow \infty$. $\square$

For the proof of Proposition A.1, we shall approximate the function $t \mapsto t^\alpha \wedge M$ with smooth functions. The next technical lemma gives sufficient conditions in order to apply a chain rule. Note that we do not require that $f'$ is bounded.

Lemma A.4

Let $f \in C^1(0,\infty )$. Suppose that f is bounded, $\lim _{t \downarrow 0} f(t) = 0$, $\lim _{t \downarrow 0} t \, f'(t) = 0$, and $\sup _{t \in (0,\infty )} t \, |f'(t)| < \infty$. Let $u \in H^1(\Omega ,H)$. Define $v = f(\Vert u\Vert _H) \, u$. Then $v \in H^1(\Omega ,H)$ and

$$\begin{aligned} \partial _k v = \left\{ \begin{array}{ll} \Vert u\Vert _H \, f'(\Vert u\Vert _H) \, {\textrm{Re}}\,({\textrm{sgn}}\,u, \partial _k u)_H \, {\textrm{sgn}}\,u + f(\Vert u\Vert _H) \, \partial _k u &{} \text{ on } [u \ne 0], \\ 0 &{} \text{ on } [u = 0], \end{array} \right. \nonumber \\ \end{aligned}$$

(5)

for all $k \in \{ 1,\ldots ,d \}$.

Proof

Let $\varepsilon > 0$. Define $v_\varepsilon = f(\sqrt{\Vert u\Vert _H^2 + \varepsilon }) \, u$. If $u \in C^1(\Omega ,H)$, then $v_\varepsilon \in C^1(\Omega ,H)$ and

$$\begin{aligned} \partial _k v_\varepsilon= & {} \sqrt{\Vert u\Vert _H^2 + \varepsilon } \, f'(\sqrt{\Vert u\Vert _H^2 + \varepsilon }) \, \frac{{\textrm{Re}}\,(u, \partial _k u)_H}{\sqrt{\Vert u\Vert _H^2 + \varepsilon }} \, \frac{1}{\sqrt{\Vert u\Vert _H^2 + \varepsilon }} \, u\nonumber \\{} & {} + f(\sqrt{\Vert u\Vert _H^2 + \varepsilon }) \, \partial _k u \end{aligned}$$

(6)

for all $k \in \{ 1,\ldots ,d \}$. Then, by Lemmas A.2 and A.3, this extends to all $u \in H^1(\Omega ,H)$ and (6) is valid. Finally choose $\varepsilon = \frac{1}{n}$, take the limit $n \rightarrow \infty$, and use again Lemma A.3. $\square$

Now we are able to prove Proposition A.1.

Proof of Proposition A.1

For all $n \in \mathbb {N}$, define $f_n,f :(0,\infty ) \rightarrow \mathbb {R}$ by

$$\begin{aligned} f(t)= & {} t^\alpha \wedge M, \\ f_n(t)= & {} {\textstyle \frac{1}{2}} \, ( t^\alpha + \sqrt{M^2 + n^{-1}} - \sqrt{|t^\alpha - M|^2 + n^{-1}} ). \end{aligned}$$

Then $\lim f_n(t) = f(t)$ for all $t \in (0,\infty )$. Also $\lim _{t \downarrow 0} f_n(t) = 0$ for all $n \in \mathbb {N}$. Let $n \in \mathbb {N}$. Then $f_n \in C^1(0,\infty )$ and

$$\begin{aligned} f_n'(t) = {\textstyle \frac{1}{2}} \, \alpha \, t^{\alpha - 1} \Big ( 1 - \frac{t^\alpha - M}{\sqrt{(t^\alpha - M)^2 + n^{-1}}} \Big ) \end{aligned}$$

for all $t \in (0,\infty )$. In particular, $f_n$ is increasing. Moreover, $\lim _{t \downarrow 0} t \, f_n'(t) = 0$. In addition, $\lim _{n \rightarrow \infty } f_n'(t) = \alpha \, t^{\alpha - 1}$ if $t^\alpha \le M$ and $\lim _{n \rightarrow \infty } f_n'(t) = 0$ if $t^\alpha > M$.

Let $n \in \mathbb {N}$ and $t \in (0,\infty )$. If $t^\alpha \le M$, then

$$\begin{aligned} 0 \le t \, f_n'(t) = {\textstyle \frac{1}{2}} \, \alpha \, t^\alpha \Big ( 1 + \frac{M - t^\alpha }{\sqrt{(t^\alpha - M)^2 + n^{-1}}} \Big ) \le \alpha \, M. \end{aligned}$$

Alternatively, if $t^\alpha > M$, then

$$\begin{aligned} 0\le & {} t \, f_n'(t) = {\textstyle \frac{1}{2}} \, \alpha \, \frac{t^\alpha - M + M}{\sqrt{(t^\alpha - M)^2 + n^{-1}}} \Big ( \sqrt{(t^\alpha - M)^2 + n^{-1}} - \sqrt{(t^\alpha - M)^2} \Big ) \\\le & {} {\textstyle \frac{1}{2}} \, \alpha \Big ( 1 + \frac{M}{\sqrt{n^{-1}}} \Big ) \sqrt{n^{-1}} \le {\textstyle \frac{1}{2}} \, \alpha \, (1 + M). \end{aligned}$$

So

$$\begin{aligned} \sup _{n \in \mathbb {N}} \sup _{t \in (0,\infty )} t \, |f_n'(t)| \le \alpha \, (M+1). \end{aligned}$$

(7)

If $n \in \mathbb {N}$ and $t \in (0,\infty )$, then

$$\begin{aligned} 0\le & {} f_n(t) \le {\textstyle \frac{1}{2}} \, (t^\alpha + M+1 - \sqrt{|t^\alpha - M|^2 + n^{-1}} )\\\le & {} {\textstyle \frac{1}{2}} \, (t^\alpha + M+1 - |t^\alpha - M| ) = {\textstyle \frac{1}{2}} + f(t) \le {\textstyle \frac{1}{2}} + M. \end{aligned}$$

So $f_n$ is bounded and even

$$\begin{aligned} \sup _{n \in \mathbb {N}} \sup _{t \in (0,\infty )} |f_n(t)| \le {\textstyle \frac{1}{2}} + M. \end{aligned}$$

(8)

Hence all conditions of Lemma A.4 are satisfied for all the $f_n$.

Let $u \in H^1(\Omega ,H)$. For all $n \in \mathbb {N}$, define $v_n = f_n(u) \, u$. Then $v_n \in H^1(\Omega ,H)$ with derivatives given by (5) and f replaced by $f_n$. The Lebesgue dominated convergence theorem and the uniform bounds (8) and (7) imply that $\lim v_n = v$ and $\lim \partial _k v_n = \partial _k v$ in $L_2(\Omega ,H)$ for all $k \in \{ 1,\ldots ,d \}$. Then the proposition follows from Lemma A.3. $\square$

Almost the same arguments show that the norm of an $H^1(\Omega ,H)$-function is in the Sobolev space.

Lemma A.5

Let $u \in H^1(\Omega ,H)$. Then $\Vert u\Vert _H \in H^1(\Omega )$ and furthermore $\partial _k \Vert u\Vert _H = {\textrm{Re}}\,({\textrm{sgn}}\,u, \partial _k u)_H$ for all $k \in \{ 1,\ldots ,d \}$.

Proof

Let $\varepsilon > 0$. For all $u \in H^1(\Omega ,H)$, define $u_\varepsilon :\Omega \rightarrow H$ by $u_\varepsilon = \sqrt{\Vert u\Vert _H^2 + \varepsilon }$. Let $\varphi \in C_c^\infty (\Omega )$ and $k \in \{ 1,\ldots ,d \}$. If $u \in C^\infty (\Omega ,H) \cap H^1(\Omega ,H)$, then $u_\varepsilon \in C^\infty (\Omega ,H)$ with classical partial derivative $\partial _k u_\varepsilon = \frac{{\textrm{Re}}\,(u, \partial _k u)_H}{u_\varepsilon }$. Hence

$$\begin{aligned} - \int \limits _\Omega u_\varepsilon \, \partial _k \varphi = \int \limits _\Omega \frac{{\textrm{Re}}\,(u, \partial _k u)_H}{u_\varepsilon } \, \varphi . \end{aligned}$$

(9)

Using approximation and Lemma A.2, it follows that (9) is valid for all $u \in H^1(\Omega ,H)$. Finally choose $\varepsilon = \frac{1}{n}$ and take the limit $n \rightarrow \infty$. $\square$

B. Strict convexity of $L_p(\Omega ,H)$

As before, let $(\Omega ,{{\mathcal {B}}},\mu )$ be a $\sigma$-finite measure space and H a Hilbert space. In order to make this paper more self-contained, we give a direct proof of the following theorem. At the end of this section, we give information on more general results.

Theorem B.1

Let $p \in (1,\infty )$ and $u,v \in L_p(\Omega ,H)$ with $\Vert u\Vert _p = \Vert v\Vert _p = 1$. If $\Vert u+v\Vert _p = 2$, then $u = v$.

For the proof of Theorem B.1, we use three lemmas.

Lemma B.2

Let $\xi ,\eta \in H$ and suppose that $\Vert \xi + \eta \Vert _H = \Vert \xi \Vert _H + \Vert \eta \Vert _H$. If $\eta \ne 0$, then there is a $\lambda \in [0,\infty )$ such that $\xi = \lambda \, \eta$.

Proof

The equality implies that ${\textrm{Re}}\,(\xi ,\eta )_H = \Vert \xi \Vert _H \, \Vert \eta \Vert _H$. This gives equality in the Cauchy–Schwarz inequality. Hence there is a $\lambda \in \mathbb {C}$ such that $\xi = \lambda \, \eta$. Using again the equality, one deduces that $|1 + \lambda | = |\lambda | + 1$ and therefore $\lambda \in [0,\infty )$. $\square$

Let $p,q \in (1,\infty )$ and suppose that $\frac{1}{p} + \frac{1}{q} = 1$.

Lemma B.3

Let $a,b \in [0,\infty )$. Then $a \, b \le \frac{1}{p} \, a^p + \frac{1}{q} \, b^q$ and the equality holds if and only if $a^p = b^q$.

Proof

This follows from the concavity of the logarithm. $\square$

Lemma B.4

Let $f \in L_p(\Omega )$ and $g \in L_q(\Omega )$ with $f \ne 0$ and $g \ne 0$. Suppose that $\int _\Omega |f| \, |g| = \Vert f\Vert _{L_p(\Omega )} \, \Vert g\Vert _{L_q(\Omega )}$. Then there exists a $\lambda > 0$ such that $|f|^p = \lambda \, |g|^q$ almost everywhere.

Proof

We may assume that $\Vert f\Vert _{L_p(\Omega )} = 1 = \Vert g\Vert _{L_q(\Omega )}$. Then

$$\begin{aligned} 1 = \int \limits _\Omega |f| \, |g| \le \int \limits _\Omega {\textstyle \frac{1}{p}} \, |f|^p + {\textstyle \frac{1}{q}} \, |g|^q = {\textstyle \frac{1}{p}} \, \Vert f\Vert _{L_p(\Omega )} + {\textstyle \frac{1}{q}} \, \Vert g\Vert _{L_q(\Omega )} = 1. \end{aligned}$$

Hence $|f| \, |g| = {\textstyle \frac{1}{p}} \, |f|^p + \frac{1}{q} \, |g|^q$ almost everywhere and the lemma follows from Lemma B.3. $\square$

Proof of Theorem B.1

Using the triangle inequality on H, twice the Hölder inequality on $L_p(\Omega )$, and the assumption $\Vert u+v\Vert _p = \Vert u\Vert _p + \Vert v\Vert _p$, one obtains

$$\begin{aligned} \Vert u + v\Vert _p^p= & {} \int \limits _\Omega \Vert u+v\Vert _H \, \Vert u+v\Vert _H^{p-1}\\\le & {} \int \limits _\Omega \Vert u\Vert _H \, \Vert u+v\Vert _H^{p-1} + \int \limits _\Omega \Vert v\Vert _H \, \Vert u+v\Vert _H^{p-1} \\\le & {} \Big ( \int \limits _\Omega \Vert u\Vert _H^p \Big )^{1/p} \Big ( \int \limits _\Omega \Vert u+v\Vert _H^{(p-1) q} \Big )^{1/q}\\{} & {} + \Big ( \int \limits _\Omega \Vert v\Vert _H^p \Big )^{1/p} \Big ( \int \limits _\Omega \Vert u+v\Vert _H^{(p-1) q} \Big )^{1/q} \\= & {} (\Vert u\Vert _p + \Vert v\Vert _p) \, \Vert u+v\Vert _p^{p/q} = \Vert u+v\Vert _p^{\frac{p}{q} + 1} = \Vert u+v\Vert _p^p. \end{aligned}$$

Hence all three inequalities are equalities. The first gives that there is a null-set $N_1 \subset \Omega$ such that $\Vert u+v\Vert _H(x) = \Vert u\Vert _H(x) + \Vert v\Vert _H(x)$ for all $x \in \Omega {\setminus } N_1$ such that $\Vert u+v\Vert _H(x) \ne 0$. Recall that $\Vert u\Vert _p = 1$, so $\Vert u\Vert _H \ne 0 \in L_p(\Omega )$. Similarly $\Vert u+v\Vert _H \ne 0 \in L_p(\Omega )$ and therefore $\Vert u+v\Vert _H^{p-1} \ne 0 \in L_q(\Omega )$. Hence the equality in the first Hölder inequality together with Lemma B.4 gives that there are $\alpha > 0$ and a null-set $N_2 \subset \Omega$ such that $\Vert u\Vert _H^p(x) = \alpha \, \Vert u+v\Vert _H^{(p-1)q}(x)$ for all $x \in \Omega {\setminus } N_2$. Similarly there are $\beta > 0$ and a null-set $N_3 \subset \Omega$ such that $\Vert v\Vert _H^p(x) = \beta \, \Vert u+v\Vert _H^{(p-1)q}(x)$ for all $x \in \Omega {\setminus } N_3$. Hence $\Vert u\Vert _H^p = \gamma \, \Vert v\Vert _H^p$ on $\Omega \setminus (N_2 \cup N_3)$, where $\gamma = \frac{\alpha }{\beta }$. Since $\Vert u\Vert _p = \Vert v\Vert _p = 1$, one deduces that $\gamma = 1$.

Now let $x \in \Omega \setminus (N_1 \cup N_2 \cup N_3)$. If $\Vert u+v\Vert _H(x) = 0$, then subsequently $\Vert u\Vert _H^p(x) = \alpha \, \Vert u+v\Vert _H^{(p-1)q}(x) = 0$ and $u(x) = 0$. Similarly $v(x) = 0$ and therefore $u(x) = v(x)$. Alternatively, if $\Vert u+v\Vert _H(x) \ne 0$, then $v(x) \ne 0$ since $x \not \in N_3$. Moreover, $\Vert u(x) + v(x)\Vert _H = \Vert u(x)\Vert _H + \Vert v(x)\Vert _H$ and Lemma B.2 implies that there is a $\lambda \in [0,\infty )$ such that $u(x) = \lambda \, v(x)$. But $\Vert u(x)\Vert _H^p = \Vert v(x)\Vert _H^p$ and hence $u(x) = v(x)$. Therefore $u = v$ almost everywhere. $\square$

With a small modification, one can prove that $L_p(\Omega ,E)$ is strictly convex if E is strictly convex and $p \in (1,\infty )$. In fact, a stronger result than Theorem B.1 is known. The space $L_p(\Omega ,E)$ is uniformly convex if E is uniformly convex and $p \in (1,\infty )$. See [10] and the references therein.