An Improved Discrete p-Hardy Inequality

Abstract

We improve the classical discrete Hardy inequality for $1<p<\infty$ for functions on the natural numbers. For integer values of p the Hardy weight is shown to have a series expansion with strictly positive coefficients. Notably, this weight is optimal, i.e. critical and null-critical.

1 Introduction and Main Result

In 1918 Hardy was looking for a simple and elegant proof of Hilbert’s theorem in the context of the convergence of double sums, [12]. Although it is not explicitly mentioned, the paper contains the essential argument for his then famous inequality. In a letter to Hardy in 1921, [20], Landau gave a proof with the sharp constant

$$\begin{aligned} \sum _{n=1}^\infty a^p_n \ge \biggl (\frac{p-1}{p} \biggr )^p \sum _{n=1}^\infty \biggl ( \frac{a_1+a_2+\ldots + a_n}{n} \biggr )^p \end{aligned}$$

for $p>1$ where $(a_n)$ is an arbitrary sequence of non-negative real numbers. This inequality was first highlighted in [13] and is referred to as a p-Hardy inequality. Since then various proofs of this inequality were given, where short and elegant ones are due to Elliott [6] and Ingham, see [13, p. 243] and by Lefèvre [21]. See also [15] for a beautiful historical survey about the origins of Hardy’s inequality.

It is not hard to see that the inequality above can be derived from the following inequality for compactly supported $\phi \in C_c(\mathbb {N})$ with $\phi (0)=0$

$$\begin{aligned} \sum _{n=1}^\infty \left|\phi (n)-\phi (n-1)\right|^p \ge \sum _{n=1}^\infty w_{p}^{H}(n){\left|\phi (n)\right|^p}, \end{aligned}$$

where

$$\begin{aligned} w_{p}^{H}(n)=\biggl (\frac{p-1}{p} \biggr )^p\frac{1}{n^p}. \end{aligned}$$

In this work, we show that the classical weight $w_p^H$ can be replaced by a pointwise strictly larger weight $w_p$, and the aforementioned Hardy inequality holds with $w_p^H$ being replaced by $w_p$. Recently, the concept of optimality was studied for general $1< p <\infty$ on weighted graphs in [7]. Employing these results, we conclude that $w_p$ is optimal, which means that

• for every function w with $w \ge w_p$ and $w \ne w_p$, the p-Hardy inequality does not hold (criticality), and in addition

• $u \notin \ell ^p(\mathbb {N}, w_p)$, where $u(n) = n^{(p-1)/p}$, $n \in \mathbb {N}$ (null-criticality).

The first optimality criterion says that there cannot be a p-Hardy weight dominating $w_p$. The second optimality criterion says that the underlying ground state u is not an eigenfunction and so, the Hardy inequality does not admit a minimizer.

Consequently, by [7, Theorem 2.6], the weight $w_p$ is also optimal at infinity, which means that for every $\lambda > 0$ and each finite set $K \subset \mathbb {N}$, the weight $(1 + \lambda )w_p$ does not yield a p-Hardy inequality for functions $\varphi$ supported outside of K. This criterion in particular shows that the constant $((p-1)/p)^p$ is optimal, a fact already known for a century, cf. [15].

We formulate the main theorem of this paper.

Theorem 1

Let $p> 1$. Then, for all $\phi \in C_c(\mathbb {N})$ with $\phi (0)=0$,

$$\begin{aligned} \sum _{n=1}^\infty \left|\phi (n)-\phi (n-1)\right|^p \ge \sum _{n=1}^\infty w_{p}(n)\left|\phi (n)\right|^p, \end{aligned}$$

where $w_{p}$ is a strictly positive function given by

$$\begin{aligned} w_p(n)=\left( 1-\left( 1-\frac{1}{n}\right) ^{\frac{p-1}{p}}\right) ^{p-1}-\left( \left( 1+\frac{1}{n}\right) ^{\frac{p-1}{p}}-1\right) ^{p-1}. \end{aligned}$$

Furthermore, $w_p$ is optimal, and we have for all $n\in \mathbb {N}$

$$\begin{aligned} w_p(n)>w_{p}^{H}(n). \end{aligned}$$

Moreover, for integer $p\ge 2$, we have $w_{p}(n)=\sum _{k\in 2\mathbb {N}_{0}}c_{k} n^{-k-p}$ with $c_{k}>0$.

Example 2

The case $p = 2$ was already covered in [16] (with optimality proven in [17]), and via a different method in [11, 19]. In this case one gets $w_2(1) = 2 - \sqrt{2}$ and for $n\ge 2$

$$\begin{aligned} w_2(n)&= \,\, - \sum _{k \in 2\mathbb {N}} \left( {\begin{array}{c}1/2\\ k\end{array}}\right) \frac{2}{n^k} \, =\,\, \frac{1}{4} \frac{1}{n^2} + \frac{5}{64} \frac{1}{n^4} + \frac{21}{512} \frac{1}{n^6} + \frac{429}{16384} \frac{1}{n^8}+ \ldots \end{aligned}$$

In the case $p=3$, one obtains $w_3(1) = \,\, 1 - (2^{2/3}-1)^2$ and for $n\ge 2$

$$\begin{aligned} w_3(n)&= \sum _{k \in 2\mathbb {N}+ 1} \left( 2\,\left( {\begin{array}{c}2/3\\ k\end{array}}\right) - \left( {\begin{array}{c}4/3\\ k\end{array}}\right) \right) \,\frac{2}{n^k} = \frac{8}{27} \frac{1}{n^3} + \frac{8}{81} \frac{1}{n^5} + \frac{112}{2187} \frac{1}{n^7} + \ldots \end{aligned}$$

In the case $p=4$, one gets $w_4(1) = \,\, 1 - (2^{3/4}-1)^3$ and for $n\ge 2$

$$\begin{aligned} {w_4(n)}&= \sum _{k \in 2\mathbb {N}+ 2} \left( 3 \left( {\begin{array}{c}3/2\\ k\end{array}}\right) -3 \left( {\begin{array}{c}3/4\\ k\end{array}}\right) - \left( {\begin{array}{c}9/4\\ k\end{array}}\right) \right) \,\frac{2}{n^k} \\&= \quad \frac{81}{256} \frac{1}{n^4} + \frac{891}{8192} \frac{1}{n^6} + \frac{58653}{1048576} \frac{1}{n^8} + \ldots \end{aligned}$$

For general $p>1$ one obtains the asymptotics

$$\begin{aligned} w_{p}(n)=\left( \frac{p - 1}{pn} \right) ^p \bigl (1+\epsilon _{p}(n)\bigr ), \end{aligned}$$

where

$$\begin{aligned} \epsilon _{p}(n) = \left( \frac{ 3}{ 8} - \frac{1}{8 p}\right) \frac{1}{n^2} + \left( \frac{215 p^3 - 38 p^2 - 31 p + 6}{1152 p^3} \right) \frac{1}{n^4} + O\left( \frac{1}{n^6}\right) . \end{aligned}$$

From this formula it is clear that $w_{p}(n)$ is strictly larger than the classical Hardy weight for large n. Note however that the theorem above states that $\epsilon _{p}(n)>0$ at all places $n \in \mathbb {N}$. It is not hard to check that $\epsilon _{p}(n)$ can be expanded into a power series with respect to 1/n where all odd coefficients vanish. Theorem 1 states that for integer $p\ge 2$ these coefficients are positive. We conjecture that all these coefficients are strictly positive for all $p>1$.

Remark 3

It is easy to see that for $1< p < \infty$, our p-Hardy inequality can be stated as follows: for every real-valued sequence $a=(a_n)$, one has

$$\begin{aligned} \sum _{n=1}^{\infty } |a_n|^p \ge \Bigg (\frac{p-1}{p}\Bigg )^p \sum _{n=1}^{\infty } \big ( 1 + \epsilon _p(n) \big ) \Bigg | \frac{1}{n} \sum _{j=1}^n a_j \Bigg |^p, \end{aligned}$$

where the function $\epsilon _p$ is as in the previous example. Denoting by C the Cesàro mean operator on $\ell ^p(\mathbb {N})$, defined as $C(a) = \frac{1}{n} \sum _{j=1}^n a_j$, the above inequality says that $C:\ell ^p(\mathbb {N}) \rightarrow \ell ^p(\mathbb {N}, \rho )$ is bounded, where $\rho = 1 + \epsilon _p$ is understood as a measure on $\mathbb {N}$. Equivalently, one obtains $\ell ^p$-boundedness of a weighted version of the Cesàro mean operator. For certain weights, such boundedness phenomena were studied recently in [22].

2 Proof of the Hardy Inequality

The combinatorial p-Laplacian $\Delta _p$ for real valued functions on $\mathbb {N}_{0}$ is given by

$$\begin{aligned} \Delta _p f(n)=\sum _{m=n\pm 1}\textrm{sgn}\left( f(n)-f(m) \right) \left|f(n)-f(m)\right|^{p-1} \end{aligned}$$

for all functions f and $n\ge 1$, where $\textrm{sgn}$ is the function which takes the value $-1$ on $(-\infty ,0)$, the value 1 on $(0,\infty )$ and 0 at 0.

The following proposition shows that the existence of a suitable positive supersolution of $\Delta _p u \ge 0$ implies the non-negativity of the corresponding energy functional. This is one of the implications of the so-called Allegretto-Piepenbrink-type theorem (see [1, 2, 23] for linear versions in the continuum, [4, 18] for a linear version in the discrete setting, [24] for a non-linear version in the continuum and [8] for a recent version in the quasi-linear discrete setting). This statement is used to show that the weight $w_p$ is in fact a p-Hardy weight.

Proposition 4

Let $p> 1$ and let $u:\mathbb {N}_0\rightarrow [0,\infty )$ be strictly positive on $\mathbb {N}$ and such that $u(0) = 0$. Suppose that $w:\mathbb {N}\rightarrow \mathbb {R}$ satisfies $\Delta _p u=wu^{p-1}$ on $\mathbb {N}$. Then for all $\phi \in C_c(\mathbb {N})$ with $\phi (0)=0$ we have

$$\begin{aligned} \sum _{n\in \mathbb {N}} \left|\phi (n)-\phi (n-1)\right|^p\ge \sum _{n\in \mathbb {N}}w(n)\left|\phi (n)\right|^p. \end{aligned}$$

The proof follows along the lines of the proof of Proposition 2.2 in [10].

Proof

Let $p> 1$. From Lemma 2.6 in [10], we obtain for all $0\le t\le 1$ and $a\in \mathbb {C}$

$$\begin{aligned} \left|a-t\right|^p \ge (1-t)^{p-1}(\left|a\right|^p-t). \end{aligned}$$

Let w be such that $\Delta _p u=wu^{p-1}$. For given $\varphi \in C_{c}(\mathbb {N})$, we can consider $\psi = \varphi / u \in C_c(\mathbb {N})$, by strict positivity of u on $\mathbb {N}$. We assume for a moment that $m,n \in \mathbb {N}$ are such that $u(n) \ge u(m)$ and $\psi (m) \ne 0$. We apply the above inequality with the choice $t=u(m)/u(n)$ and $a=\psi (n)/\psi (m)$ in order to obtain

$$\begin{aligned} \big | (u\psi )(n) - (u\psi )(m) \big |^p \ge \big |u(n) - u(m) \big |^{p-1} \big ( |\psi (n)|^p u(n) - |\psi (m)|^p u(m) \big ). \end{aligned}$$

Further, since $u^{p}(n) \ge |u(n) - u(m)|^{p-1} u(n)$, the above inequality remains true even if $\psi (m) = 0$. Summing over $\mathbb {N}$, we obtain

$$\begin{aligned}&{\sum _{n\in \mathbb {N}} \left|(u\psi )(n)-(u\psi )(n-1)\right|^p}\\&\ge \sum _{n\in \mathbb {N}}\textrm{sgn}(u(n)-u(n-1))\left|u(n)-u(n-1)\right|^{p-1}\cdot \\&\qquad \bigl (\left|\psi (n)\right|^pu(n)-\left|\psi (n-1)\right|^pu(n-1) \bigr ) \\&=\sum _{n\in \mathbb {N}} \Delta _p u(n)\left|\psi (n)\right|^pu(n) \, = \, \sum _{n\in \mathbb {N}} w(n) u^p(n)\left|\psi (n)\right|^p. \end{aligned}$$

Note that the latter two equalities follow from rearranging the involved sums while recalling that $u(0)= 0$, and using the assumption $\Delta _p u=wu^{p-1}$. Recalling that $\phi =u\psi$, we infer the statement. $\square$

Next we show that for the weight $w_{p}$ on $\mathbb {N}$ taken from Theorem 1

$$\begin{aligned} w_p(n)=\left( 1-\left( 1-{1}/{n}\right) ^{{(p-1)}/{p}}\right) ^{p-1}-\left( \left( 1+{1}/{n}\right) ^{{(p-1)}/{p}}-1\right) ^{p-1}, \end{aligned}$$

there is a suitable positive function u such that $\Delta _{p}u =w_{p}u^{p-1}$.

Proposition 5

Let $p> 1$. Then, the function $u:\mathbb {N}_0 \rightarrow [0,\infty )$, $u(n)=n^{(p-1)/p}$ satisfies

$$\begin{aligned} \Delta _{p}u=w_{p}u^{p-1}\qquad \text{ on } \mathbb {N}. \end{aligned}$$

Proof

One directly checks that for all $n\in \mathbb {N}$

$$\begin{aligned} \frac{\Delta _{p}u(n)}{u^{p-1}(n)} =\frac{\Delta _p n^{(p-1)/p}}{n^{(p-1)^{2}/p}}=w_{p}(n) \end{aligned}$$

which immediately yields the statement. $\square$

The choice of the function u in the previous proposition is motivated by the so-called supersolution construction which yields optimal p-Hardy weights both in the continuum case for $p> 1$, cf. [3, 5], as well as for graphs with $p=2$, cf. [17]. Moreover, for $p=2$, the function $u(n) = n^{1/2}$ arises naturally in the method applied in [11, 19] for a proof of the optimal Hardy inequality on the line graph.

Combining the two propositions above already yields the p-Hardy inequality with the weight $w_{p}$. Next we show that $w_{p}$ is strictly larger than the classical Hardy weight $w_{p}^{H}(n)=\bigl ({(p-1)}/{p }\bigr )^{p}{n^{-p}}$ for all $n\in \mathbb {N}$.

3 Proof of $w_p>w_{p}^{H}$

In this section we show that the weight

$$\begin{aligned} w_p(n)=\left( 1-\left( 1-\frac{1}{n}\right) ^{\frac{p-1}{p}}\right) ^{p-1}-\left( \left( 1+\frac{1}{n}\right) ^{\frac{p-1}{p}}-1\right) ^{p-1} \end{aligned}$$

from the main theorem, Theorem 1, is strictly larger than the classical p-Hardy weight

$$\begin{aligned} w^H_p(n)=\left( \frac{p-1}{p} \right) ^p\frac{1}{n^p}. \end{aligned}$$

In fact, for fixed $p\in (1,\infty )$, we analyze the function $w:[0,1]\rightarrow [0,\infty )$

$$\begin{aligned} w(x)&=\left( 1-(1-x)^{1/q}\right) ^{p-1}-\left( (1+x)^{1/q}-1\right) ^{p-1} \end{aligned}$$

for $x \in [0,1/2]$ and $x=1$, where $q\in (1,\infty )$ is such that $1/p + 1/q=1$. Specifically, we show

$$\begin{aligned} w(x)> \left( \frac{x}{q}\right) ^{p}. \end{aligned}$$

The case $x=1$ is simple and is treated at the end of the section. The proof for $x\le 1/2$ is also elementary but more involved. We proceed by bringing $w_{p}$ into form for which we then analyze its parts. This will be eventually done by a case distinction depending on p.

Recall the binomial theorem for $r \in [0,\infty )$ and $0\le x \le 1$

$$\begin{aligned} (1\pm x)^{r}= \sum _{k=0}^\infty \left( {\begin{array}{c}r\\ k\end{array}}\right) (\pm 1)^kx^k \end{aligned}$$

where $\left( {\begin{array}{c}r\\ 0\end{array}}\right) =1, \left( {\begin{array}{c}r\\ 1\end{array}}\right) =r$ and $\left( {\begin{array}{c}r\\ k\end{array}}\right) =r(r-1)\cdots (r-k+1)/k!$ for $k\ge 2$ which is derived from the Taylor expansion of the function $x\mapsto (1\pm x)^{r}$. Applying this formula to the function w from above we obtain

$$\begin{aligned} w(x)&=\left( -\sum _{k=1}^{\infty }\left( {\begin{array}{c}{1/q }\\ k\end{array}}\right) (-x)^{k}\right) ^{p-1}-\left( \sum _{k=1}^{\infty }\left( {\begin{array}{c}{1/q }\\ k\end{array}}\right) x^{k}\right) ^{p-1}\\&=\left( \frac{x}{q}\right) ^{p-1}\left( \left( q\sum _{k=0}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) (-x)^{k}\right) ^{p-1}-\left( q\sum _{k=0}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k}\right) ^{p-1}\right) \end{aligned}$$

To streamline notation we set

$$\begin{aligned} g(x)=q\sum _{k=1}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k}. \end{aligned}$$

Note that since $q\left( {\begin{array}{c}1/q\\ 1\end{array}}\right) =1$ and $q\left|\left( {\begin{array}{c}1/q\\ k\end{array}}\right) \right|<1$ for $k\ge 2$, we have ${0< |g(\pm x)|<1}$ for $0< x\le 1/2$. Thus, we can apply the binomial theorem to $\bigl (1+g(\pm x)\bigr )^{p-1}$ in order to get

$$\begin{aligned} w(x)&=\left( \frac{x}{q}\right) ^{p-1}\Bigl (\bigl (1+g(-x)\bigr )^{p-1} - \bigl (1+g(x)\bigr )^{p-1}\Bigr )\\&=\left( \frac{x}{q}\right) ^{p-1} \left( \sum _{n=0}^\infty \left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )\right) \\&=\left( \frac{x}{q}\right) ^{p-1} \left( \left( {\begin{array}{c}p-1\\ 1\end{array}}\right) \bigl (g(-x)-g(x)\bigr ) +\sum _{n=2}^{\infty }\left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )\right) \end{aligned}$$

Thus, we have to show that the second factor on the right hand side is strictly larger than x/q. Using $q=p/(p-1)$ we compute the first term in the parenthesis on the left hand side

$$\begin{aligned} \left( {\begin{array}{c}p-1\\ 1\end{array}}\right)&\bigl (g(-x)-g(x)\bigr )=q(p-1)\sum _{k=1}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) \left( (-x)^{k} - x^{k}\right) \\&=\frac{q(p-1)(1/q)(1/q-1)}{2}(-2x)+q(p-1)\sum _{k=2}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) \left( (-x)^{k} - x^{k}\right) \\&=\frac{x}{q}-2p\sum _{k\in 2\mathbb {N}+1}\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k}\\&=\frac{x}{q}+E_{p}(x), \end{aligned}$$

with

$$\begin{aligned} E_p(x) = -2p\sum _{k\in 2\mathbb {N}+1}\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k} > 0 \end{aligned}$$

since $-2p \left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) > 0$ for odd k and $x > 0$. So, it remains to show that for the term

$$\begin{aligned} F_{p}(x)=\sum _{n=2}^{\infty }\left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr ) \end{aligned}$$

we have for $0<x\le 1/2$

$$\begin{aligned} E_{p}(x)+F_{p}(x)> 0. \end{aligned}$$

Specifically, we then get with the substitution $x=1/n$

$$\begin{aligned} w_{p}(n)=w(1/n)=\left( \frac{1}{nq}\right) ^{p-1} \left( \frac{1}{nq}+E_{p}(1/n)+F_{p}(1/n)\right) >\frac{1}{(nq)^p}=w_{p}^{H}(n) \end{aligned}$$

for $n\ge 2$.

Remark 6

It is not hard to see that $F_{p}\ge 0$ whenever $p\in \mathbb {N}$ is integer valued. Indeed, $g(-x)\ge g(x)$ as all terms in the sum $g(-x)$ are positive since $- \left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) \ge 0$ for odd k, while the terms in g(x) alternate, (they are positive for even k and negative for odd k). Moreover for positive integers p the binomial coefficients $\left( {\begin{array}{c}p-1\\ n\end{array}}\right)$ are positive. Thus, the Hardy weight we computed is larger than the classical one for integer p.

Let us now turn to the proof of

$$\begin{aligned} E_{p}(x)+F_{p}(x)>0 \end{aligned}$$

for $p \in (1,\infty )$ and $0< x\le 1/2$.

We collect the following basic properties of the function g which were partially already discussed above and will be used subsequently.

Lemma 7

For $p\in (1,\infty )$ and $0<x\le 1/2$, we have

$$\begin{aligned} -1<g(x)< 0< -g(x)< g(-x)<1. \end{aligned}$$

Proof

The function g is given by $g(x)=q\sum _{k=1}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k}$. Since $q>1$, the coefficients $b_k=q\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right)$ are negative for odd k and positive for even k. Furthermore, the sequence $(|b_{k} |)$ takes values strictly less than 1 and decays monotonically. Thus, the asserted inequalities follow easily. $\square$

We distinguish the following three cases depending on p for which the arguments are quite different:

• p lies between an odd and an even number with the subcases:

  • $\bullet$ $p\in [3,\infty )$

  • $\bullet$ $p\in (1,2]$

• p lies between an even and an odd number.

Here, for $a,b \in \mathbb {N}$, we say that p is between a and b if $a \le p \le b$.

We start with investigating the case of p lying between an odd and an even number. To this end we consider two subsequent summands as they appear in the sum given by $F_{p}$ and show that they are positive. (Indeed the sum in $F_{p}$ starts at $n=2$ but we also consider the corresponding term for $n=1$.)

Lemma 8

Let p be between an odd and an even integer. Then, for all $0< x\le 1/2$ and odd $n\in 2\mathbb {N}-1$

$$\begin{aligned} \left( {\begin{array}{c}{p-1 }\\ n\end{array}}\right)&\bigl (g^{n}(-x)-g^{n}(x)\bigr )+\left( {\begin{array}{c}{p-1 }\\ n+1\end{array}}\right) \bigl (g^{n+1}(-x)-g^{n+1}(x)\bigr )\ge 0. \end{aligned}$$

Proof

Let p be between an odd and an even integer. We first consider $n < p-1$. In the case $k\le p-1$, we have $\left( {\begin{array}{c}{p-1 }\\ k\end{array}}\right) \ge 0$. So, the statement for $n < p-1$ follows directly from Lemma 7 as $|g(x)|< 1$ for $0< x\le 1/2$. (Observe that $n+1\le p-1$ for $n < p-1$ and $n\in 2\mathbb {N}-1$ as p is between an odd and an even integer.)

On the other hand, for odd $n\in 2\mathbb {N}-1$ with $n\ge p-1$,

$$\begin{aligned} \left( {\begin{array}{c}{p-1 }\\ n\end{array}}\right) \ge -\left( {\begin{array}{c}{p-1 }\\ n+1\end{array}}\right) \ge 0. \end{aligned}$$

From Lemma 7 we know that $g^{n+1}(x)\ge 0 \ge g^n(x)$ for odd $n\in 2\mathbb {N}-1$ and $0\le x \le 1/2$.

We obtain

$$\begin{aligned} \left( {\begin{array}{c}{p-1 }\\ n\end{array}}\right)&\bigl (g^{n}(-x)-g^{n}(x)\bigr )+\left( {\begin{array}{c}{p-1 }\\ n+1\end{array}}\right) \bigl (g^{n+1}(-x)-g^{n+1}(x)\bigr )\\&=\left( {\begin{array}{c}{p-1 }\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )-\left|\left( {\begin{array}{c}{p-1 }\\ n+1\end{array}}\right) \right|\bigl (g^{n+1}(-x)-g^{n+1}(x)\bigr )\\&\ge \left( {\begin{array}{c}{p-1 }\\ n\end{array}}\right) g^{n}(-x)-\left|\left( {\begin{array}{c}{p-1 }\\ n+1\end{array}}\right) \right|g^{n+1}(-x)\\&\ge \left|\left( {\begin{array}{c}{p-1 }\\ n+1\end{array}}\right) \right|\left( g^{n}(-x)-g^{n+1}(-x)\right) \\&\ge 0, \end{aligned}$$

where the last inequality follows from $0\le g(-x)< 1$ for $0\le x\le 1/2$ thanks to Lemma 7. $\square$

With Lemma 8 we can treat the case of $p\ge 3$ lying between an odd and an even number. This is done in the next proposition.

Proposition 9

Let $p \ge 3$ be between an odd and an even integer. Then, for all $0< x\le 1/2$ we have $F_{p}(x)\ge 0$ and

$$\begin{aligned} E_{p}(x)+F_{p}(x)>0. \end{aligned}$$

In particular, $w_{p}(n)>w_{p}^{H}(n)$ for $n\ge 2$.

Proof

We can write $F_{p}(x)=\sum _{n=2}^{\infty }\left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )$ as

$$\begin{aligned}&F_{p}(x)=\left( {\begin{array}{c}p-1\\ 2\end{array}}\right) \bigl (g^{2}(-x)-g^{2}(x)\bigr )\\&+\sum _{n\in 2\mathbb {N}+1}^{\infty } \left( \left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr ) + \left( {\begin{array}{c}p-1\\ n+1\end{array}}\right) \bigl (g^{n+1}(-x)-g^{n+1}(x)\bigr )\right) \end{aligned}$$

By Lemma 8 the terms in the sum on the right hand side are all positive. Furthermore, $\left( {\begin{array}{c}p-1\\ 2\end{array}}\right) \ge 0$ for $p\ge 3$ and $g(-x)\ge |g(x) |$ by Lemma 7. Thus, also the first term on the right hand side is positive as well and $F_{p}\ge 0$ follows. From the discussion in the beginning in the section we take $E_{p}(x)>0$ for $0<x \le 1/2$. The ”in particular“ follows from the discussion above Lemma 7.

$\square$

Note that we cannot treat the case $1\le p\le 2$ in the same way since the sum in $F_{p}$ starts at the index $n=2$. Hence, there is still a negative term $\left( {\begin{array}{c}p-1\\ 2\end{array}}\right) \bigl (g^{2}(-x)-g^{2}(x)\bigr ).$ We deal with this case, $1\le p\le 2$, next.

We denote the Taylor coefficients of $x\mapsto g(-x)$ by $a_{k}$, i.e.,

$$\begin{aligned} g(-x)&=q\sum _{k=1}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) (-x)^{k} = \sum _{k=1}^{\infty }a_{k}x^k,\\ g(x)&=q\sum _{k=1}^{\infty }\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k} = \sum _{k=1}^{\infty }a_{k}(-1)^kx^k. \end{aligned}$$

The function $E_{p}(x)=-2p\sum _{k\in 2\mathbb {N}+1}\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k}$ is odd and, therefore, we have

$$\begin{aligned} E_{p}(x)=2(p-1)\sum _{n=1}^{\infty }a_{2n+1}x^{2n+1}. \end{aligned}$$

Furthermore, recall that $E_p(x) > 0$ for $x > 0$, since $-2p \left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) > 0$ for odd k.

Lemma 10

Let ${p\ge 1}$ and ${0\le x\le 1/2}$. Then,

$$\begin{aligned} g(-x)+g(x)\le \frac{4}{9}\cdot \frac{(p+1)}{p^2} x^2. \end{aligned}$$

Proof

We calculate using $a_{2}\ge a_{n}$ for $n\ge 2$, the geometric series, $x\le 1/2$ and the specific value of the Taylor coefficient $a_2=q\left( {\begin{array}{c}1/q\\ 3\end{array}}\right) =\frac{(p+1)}{6p^2}$

$$\begin{aligned} g(-x)+g(x)&=2\sum _{k=1}^\infty a_{2k}x^{2k}\le 2a_2\frac{x^2}{1-x^2}\le \frac{8}{3}a_2 x^2 =\frac{4}{9}\cdot \frac{(p+1)}{p^2}x^2. \end{aligned}$$

$\square$

With the help of this lemma and Lemma 8 we can treat the case $p\in (1,2]$.

Proposition 11

Let $p \in (1,2]$. Then, for all $0< x\le 1/2$, we have

$$\begin{aligned} E_{p}(x)+F_{p}(x)>0. \end{aligned}$$

In particular, $w_{p}(n)>w_{p}^{H}(n)$ for $n\ge 2$.

Proof

We show $E_{p}+F_{p}>0$ and deduce the “in particular” from the discussion above Lemma 7. By Lemma 8 we have for all $0< x\le 1/2$

$$\begin{aligned} F_{p}(x) =&\left( {\begin{array}{c}p-1\\ 2\end{array}}\right) \bigl (g^{2}(-x)-g^{2}(x)\bigr )+\sum _{n\in 2\mathbb {N}+1}^{\infty } \left( \left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )\right. \\&\left. + \left( {\begin{array}{c}p-1\\ n+1\end{array}}\right) \bigl (g^{n+1}(-x)-g^{n+1}(x)\bigr )\right) \\ \ge&\left( {\begin{array}{c}p-1\\ 2\end{array}}\right) \bigl (g^{2}(-x)-g^{2}(x) \bigr )\\ =&\frac{p-2}{2}\bigl (g(-x)+g(x) \bigr )\left( E_{p}(x)+\frac{p-1}{p}x \right) \\ \ge&\frac{2}{9}\cdot \frac{(p-2)(p+1)}{p^2}\left( E_{p}(x)+\frac{p-1}{p}x \right) \cdot x^2\\ \ge&- \frac{1}{9}E_{p}(x)+\frac{2}{9}\cdot \frac{(p-2)(p-1)(p+1)}{p^3}\cdot x^3 \end{aligned}$$

where we used the definition of $E_{p}$, i.e., $(p-1)\bigl (g(-x)-g(x)\bigr )=E_{p}(x)+ \frac{p-1}{p}x$ and Lemma 10 which is justified since $E_p(x)>0$ and $p-2<0$. Moreover, in the last step we estimated the coefficient of the first term in its minimum in $p=1$ and $x=1/2$.

Now, we use the representation of $E_{p}$ as a power series to estimate

$$\begin{aligned} E_p(x) = -2p\sum _{k\in 2\mathbb {N}+1}\left( {\begin{array}{c}{1/q }\\ k+1\end{array}}\right) x^{k} \ge -2p_{}\left( {\begin{array}{c}{1/q }\\ 4\end{array}}\right) x^{3}=\frac{(p-1)(p+1)(2p+1)}{12p^3}x^3. \end{aligned}$$

Putting this together with the estimate on $F_{p}$ above, we arrive at

$$\begin{aligned} E_{p}(x)+F_{p}(x)&\ge \frac{8}{9} E_{p}(x) +\frac{2}{9}\cdot \frac{(p-2)(p-1)(p+1)}{p^3}\cdot x^3\\&\ge \left( \frac{8}{9} \frac{(p-1)(p+1)(2p+1)}{12p^3} +\frac{2}{9}\cdot \frac{(p-2)(p-1)(p+1)}{p^3}\right) \cdot x^3\\&=\frac{10(p-1)^2(p+1)}{27p^3}\cdot x^3 \end{aligned}$$

Hence, it remains to consider the case of p between an even and an odd integer for which we need the following three lemmas.

Lemma 12

Let $p, q \ge 1$ such that $1/p+1/q=1$ and $k\ge 2$. Then,

$$\begin{aligned} a_{k}=q\left| \left( {\begin{array}{c}1/q\\ k+1\end{array}}\right) \right| \ge \frac{1}{pk(k+1)}=\frac{1}{q(p-1)k(k+1)}. \end{aligned}$$

Proof

We calculate using $1/p+1/q=1$

$$\begin{aligned} q\left| \left( {\begin{array}{c}1/q\\ k+1\end{array}}\right) \right|&=\frac{(1-1/q)(2-1/q)(3-1/q)\cdots (k-1/q)}{(k+1)!} \\&=\frac{1}{pk(k+1)} \frac{(1+1/p)(2+1/p)\cdots ((k-1)+1/p)}{(k-1)!} \\&= \frac{1}{pk(k+1)} \left( 1+\frac{1}{p}\right) \left( 1+\frac{1}{2p}\right) \cdots \left( 1+\frac{1}{(k-1)p}\right) \\&\ge \frac{1}{pk(k+1)}. \end{aligned}$$

Lemma 13

Let $p, q\in (1,\infty )$ such that $1/p+1/q=1$ and $k\in \mathbb {N}, k> p$. Then,

$$\begin{aligned} \left| \left( {\begin{array}{c}p-1\\ k\end{array}}\right) \right| \le \frac{1}{4(p-1)}=\frac{(q-1)}{4}. \end{aligned}$$

Proof

Let $n\in \mathbb {N}$ be such that $n-1\le p\le n$. Moreover, let $\gamma = p-(n-1)$, i.e., $1-\gamma =n-p$, so, $\gamma \in [0,1]$. Since $k>p$ and $n,k\in \mathbb {N}$, we have that $k\ge n$ and therefore,

$$\begin{aligned} \left| \left( {\begin{array}{c}p-1\\ k\end{array}}\right) \right|&= \left| \frac{(p-1)(p-2)\cdots (p-(n-1)) (p-n)\cdots (p-k)}{k!}\right| \\&=\left| {\left( \frac{p-1}{n-1}\right) \left( \frac{p-2}{n-2}\right) \cdots \frac{\bigl ({p}-(n-1)\bigr ) }{1}\left( \frac{p-n}{n}\right) \cdots \left( \frac{p-k}{k}\right) }\right| \\&\le \left| \frac{({p}-(n-1))(p-n)}{n}\right| =\frac{\gamma (1-\gamma )}{n} \le \frac{1}{4(p-1)}=\frac{(q-1)}{4}. \end{aligned}$$

Lemma 14

For $0< x\le 1/2$ and $q>1$, we get

$$\begin{aligned} g(-x)\le \frac{(q-1)(5q-1)}{6q^2}x. \end{aligned}$$

Proof

We calculate using $a_2\ge a_k$ for $k\ge 2$

$$\begin{aligned} g(-x)&= q\left( \left| \left( {\begin{array}{c}{1/q }\\ 2\end{array}}\right) \right| + \sum _{k=1}^{\infty }\left| \left( {\begin{array}{c}{1/q }\\ k+2\end{array}}\right) \right| x^{k}\right) x\\&\le q\left( \left| \left( {\begin{array}{c}{1/q }\\ 2\end{array}}\right) \right| + \sum _{k=1}^{\infty }\left| \left( {\begin{array}{c}{1/q }\\ k+2\end{array}}\right) \right| 2^{-k}\right) x\\&\le q\left( \left| \left( {\begin{array}{c}{1/q }\\ 2\end{array}}\right) \right| + \left| \left( {\begin{array}{c}{1/q }\\ 3\end{array}}\right) \right| \right) x\\&=\frac{(q-1)(5q-1)}{6q^2}x. \end{aligned}$$

With the help of these lemmas we can finally treat the case where p lies between an even and an odd number.

Proposition 15

Let $p \in [2,\infty )$ be between an even and an odd integer. Then, for all $0< x\le 1/2$ we have

$$\begin{aligned} E_{p}(x)+F_{p}(x)>0. \end{aligned}$$

In particular, $w_{p}(n)>w_{p}^{H}(n)$ for $n\ge 2$.

Proof

Clearly, we have $\left( {\begin{array}{c}{p-1 }\\ n\end{array}}\right) \ge 0$ for $n\le p$ and for $n\in 2\mathbb {N}$. Since we have $g(-x)\ge |g(x)|$ by Lemma 7, we obtain for the first $n\le p$ terms and the terms for even n in $F_{p}(x)$ that

$$\begin{aligned} \left( {\begin{array}{c}{p-1 }\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )\ge 0. \end{aligned}$$

Note that $E_{p}(x)= 2(p-1)\sum _{n=1}^{\infty }a_{2n+1}x^{2n+1}> 2(p-1)\sum _{n=k}^{\infty }a_{2n+1}x^{2n+1}$ since the coefficients $a_k$ are positive. With the observation made at the beginning of the proof, this leads to

$$\begin{aligned} E_{p}(x)+F_{p}(x)> \sum _{n\in 2\mathbb {N}+1, n\ge p}\left( 2(p-1)a_{n}x^{n}+\left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )\right) . \end{aligned}$$

We continue to show that all the terms in the sum are strictly positive which finishes the proof. To this end note that for $n\ge p$ with $n\in 2\mathbb {N}+1$, we use $g(-x) \ge |g(x)|$, see Lemma 7, as well as $\left( {\begin{array}{c}p-1\\ n\end{array}}\right) \le 0$ in the first step and the estimate on $\left( {\begin{array}{c}p-1\\ n\end{array}}\right)$, see Lemma 13, and the estimate on $g(-x)$, see Lemma 14, in the second step in order to get

$$\begin{aligned} \left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )&\ge 2\left( {\begin{array}{c}p-1\\ n\end{array}}\right) g^{n}(-x)\\ {}&\ge -\frac{(q-1)}{2}\left( \frac{(q-1)(5q-1)}{6q^2}\right) ^{n}x^{n} \end{aligned}$$

We use the estimate on $a_{n}$, Lemma 12,

$$\begin{aligned} 2(p-1)a_{n} x^{n} \ge 2\frac{1}{qn(n+1)}x^{n}. \end{aligned}$$

Next, we put these two estimates together and find that the minimum in the coefficient is clearly assumed at $q=2$ since $p\ge 2\ge q$

$$\begin{aligned}{} & {} \left( 2(p-1)a_{n}x^{n}+\left( {\begin{array}{c}p-1\\ n\end{array}}\right) \bigl (g^{n}(-x)-g^{n}(x)\bigr )\right) \\{} & {} \qquad \ge 2\left( \frac{1}{qn(n+1)}-\frac{(q-1)}{4}\left( \frac{(q-1)(5q-1)}{6q^2}\right) ^{n}\right) x^{n}\\{} & {} \qquad \ge \left( \frac{1}{n(n+1)}-\frac{1}{2}\left( \frac{3}{8}\right) ^{n}\right) x^{n}\ge \left( \frac{1}{n(n+1)} -\frac{1}{2^{n+1}}\right) x^{n}>0, \end{aligned}$$

where the positivity follows by a simple induction argument. This concludes the proof by noticing that the “in particular” part follows from the discussion above Lemma 7. $\square$

In summary, the above considerations yield

$$\begin{aligned} E_{p}(x)+F_{p}(x)>0 \end{aligned}$$

for $p\in (1,\infty )$ and $0< x\le 1/2$. By the discussion at the beginning of the section this yields $w_{p}(n)>w_{p}^{H}(n)$ for $n\ge 2$.

We finish the section by treating the case $n=1$ which corresponds to $x=1$. With this we finally conclude that $w_{p}(n)>w_{p}^{H}(n)$ for all $n\ge 1$ in the next section.

Proposition 16

Let $p\in (1,\infty )$. Then, $w_p(1)> w^H_p(1)$.

Proof

Recall that $w_p(1) =1-(2^{1-1/p}-1)^{p-1}$ and $w_{p}^{H}=(1-1/p)^{p}$. By the mean value theorem applied to the function $[1,2]\rightarrow [1,2^{1-1/p}]$, $t\mapsto t^{1-1/p}$ we find

$$\begin{aligned} 2^{1-1/p}-1<1-\frac{1}{p}. \end{aligned}$$

Therefore,

$$\begin{aligned} w_{p}(1)-w_{p}^{H}(1)>1-\left( 1-\frac{1}{p}\right) ^{p-1}-\left( 1-\frac{1}{p}\right) ^{p}. \end{aligned}$$

Now the function $\psi :(1,\infty ) \rightarrow (0,\infty )$, $p\mapsto \left( 1-{1}/{p}\right) ^{p-1}+\left( 1-{1}/{p}\right) ^{p}$ is strictly monotonically decreasing because

$$\begin{aligned} \psi '(p)= \frac{1}{p-1}\left( \frac{p - 1}{p}\right) ^p \left( (2 p - 1) \log \left( \frac{p - 1}{p}\right) + 2\right) <0, \end{aligned}$$

since $\theta :p\mapsto (2p-1)\log (p-1)/p$ is strictly monotonically increasing and we have $\lim _{p\rightarrow \infty }\theta (p) = -2$. Hence, we conclude

$$\begin{aligned} w_{p}(1)-w_{p}^{H}(1)> 1-\psi (p) > 1-\lim _{t\rightarrow 1}\psi (t)=0. \end{aligned}$$

4 Proof of Theorem 1

Proof of Theorem 1

Combining Propositions 4 and 5 from Sect. 2 yields that $w_{p}$ satisfies the Hardy inequality. In Sect. 3, one obtains from Proposition 9, Proposition 11, Proposition 15 and Proposition 16 that $w_{p}>w_{p}^{H}$ on $\mathbb {N}$.

The optimality of the p-Hardy weight, i.e., for every $w\gneq w_p$ the p-Hardy inequality fails (criticality), and in addition $u \not \in \ell ^p(\mathbb {N},w_p)$, where $u(n)=n^{(p-1)/p}, n\in \mathbb {N}$ (null-criticality) can be deduced from [7, Theorem 2.3]. To this end one has to check that the function $v> 0$ which gives rise to the Hardy weight via the supersolution construction, i.e., $u=v^{(p-1)/p}$ and $w_p = \Delta _{p}u / u^{(p-1)}$ is proper, i.e., the preimage of every compact set in $(0,\infty )$ is compact, and of bounded oscillation, i.e., $C^{-1}\le v(n)/v(n+1)\le C$ for some $C>0$ and all n. These two properties are clearly satisfied for the identity function $\textrm{id}(n)=n$ which was used in Proposition 5.

Note that given criticality, null-criticality is also a simple consequence of $w_p> w_p^H$ on $\mathbb {N}$ and since $u(n)=n^{(p-1)/p}$ is not in $\ell ^{p}( \mathbb {N}, w^H_{p})$. Moreover, in the remark below, Remark 17, we give a sketch of the proof of criticality to give the reader an idea of the arguments involved.

To see the statement about the coefficients in the series expansion of $w_{p}$ for integer $p \ge 2$, recall the function

$$\begin{aligned} w(x)=\bigl (1-(1-x)^{1/q}\bigr )^{p-1}-\bigl ((1+x)^{1/q}-1\bigr )^{p-1} \end{aligned}$$

with $1/p+1/q=1$ on [0, 1] from the previous section. It is easy to check that the function $w_{+}:x\mapsto {\bigl (1-(1-x)^{1/q}\bigr )^{p-1}}$ is absolutely monotonic on [0, 1) for integer $p\ge 2$ with strictly positive derivatives. On the other hand, expanding the function $w_{-}:x\mapsto {\bigl ((1+x)^{1/q}-1\bigr )^{p-1}}$ at $x=0$, we observe that it has the same Taylor coefficients in absolute value as $w_{+}$. However, the signs alternate such that for the difference $w_{+}-w_{-}$ of these two functions the even/odd coefficients cancel for odd/even p. Furthermore, in the Taylor expansion of w at $x=0$ the first non-zero coefficient is the one for $x^{p}$ (confer Sect. 3). $\square$

Remark 17

(Sketch of the proof of criticality) Criticality of the weight $w_{p}$ is equivalent to existence of a null sequence, i.e., existence of $0\le \varphi _N\in C_{c} (\mathbb {N})$ with $\varphi (0)=0$, $\varphi (1)=1$ and $\varphi _{N}$ converging pointwise to $u(n)=n^{1/q}$ and

$$\begin{aligned} h(\varphi _{N})\!:=\!\sum _{n=0}^{\infty }|\varphi _{N}(n)-\varphi _{N}(n+1)|^{p} \!-\!\sum _{n=1}^{\infty }w_{p}^{H}(n)|\varphi _{N}(n)|^{p} \!\rightarrow \! 0,\quad N\rightarrow \infty \end{aligned}$$

cf. [8, Theorem 5.1]. We denote $a\vee b =\max \{a,b\}$ for $a,b\in \mathbb {R}$ and choose

$$\begin{aligned} \varphi _N=u\psi _{N}\qquad \text{ with } \qquad \psi _{N}(n)= 0\vee \left( 1-\frac{\log n}{\log N}\right) \end{aligned}$$

for $n,N\in \mathbb {N}$ and $u(n)=n^{1/q}$. We now use the simplified energy from [9, Theorem 3.1] which serves as a substitute of a ground state transform for $p\ne 2$

$$\begin{aligned} h(u\psi ) \asymp h_{u}(\psi ):= & {} \sum _{n=0}^{\infty }u(n)u(n+1)|\psi (n)-\psi (n+1)|^2 \cdot \\{} & {} \times \Bigg ((u(n)u(n+1))^{\frac{1}{2}}|\psi (n)-\psi (n+1)|\\{} & {} \qquad \quad +\frac{|\psi (n)|+|\psi (n+1)|}{2}|u(n)-u(n+1)|\Bigg )^{p-2}, \end{aligned}$$

where $\asymp$ means that there are two sided estimates with positive constants independent of $\psi$. We employ $u(n)=n^{1/q}$ and the definition of $\varphi _{N}, \psi _{N}$ to obtain

$$\begin{aligned} h(\varphi _N) \asymp h_{u}(\psi _{N})\le \frac{C}{\log ^{p}N} \left( \sum _{n=1}^{N}n^{p-1}\log ^{p}\left( 1+\frac{1}{n}\right) +\sum _{n=1}^{N}n^{-1}\log ^{p-2}\left( \frac{N}{n}\right) \right) , \end{aligned}$$

where we used $a(b+c)^{r}\le C( ab^{r}+ac^{r})$ for all $a,b,c\ge 0$, $r\in \mathbb {R}$ and some $C=C(r)$ to split the sum into two sums, $|(0\vee a)-(0\vee b)|\le |a-b|$ for all $a,b\in \mathbb {R}$ and $|n^r-(n+1)^r|\asymp n^{r-1}$ for $r\in (0,\infty )$ cf. e.g. [14, Lemma 2.28]. Now, using $\log (1+1/n)\le 1/n$ and again $a(b+c)^{r}\le C( ab^{r}+ac^{r})$, we infer

$$\begin{aligned} h_{u}(\psi _{N})\le \frac{C}{\log ^{p}N}\left( \log N +\log ^{p-1}N\right) \rightarrow 0,\qquad N\rightarrow \infty \end{aligned}$$

which finishes the proof.

While we know that all Taylor coefficients of the Hardy weight $w_{p}$ are strictly positive for integer $p\ge 2$, we only know $w_{p}>w^{H}_p$ for non-integer $p>1$. This leads us to the following conjecture.

Conjecture

We conjecture that $x\mapsto w(x)$ as defined in Sect. 3 is absolutely monotonic.