Introduction to Class of Uniformly Fractional Differentiable Functions

Abstract

In this paper, authors introduced new concept of uniformly fractional differentiable functions on an arbitrary interval I of R by using Caputo-type fractional derivative instead of the commonly used first-order derivative. Their interesting properties with few illustrations have been discussed in this paper.

1 Introduction

The fractional calculus is a theory of integrals and derivatives of arbitrary order, which unify and generalize the notions of integer-order differentiation and n-fold integration. We shall explain the result connected to classical analysis, namely uniformly differential functions given by Patel [1], can be extended to fractional calculus, i.e they can be generalized by replacing the first order the first derivatives and integrals, respectively, by derivatives and integrals of non-integer. The uniformly differentiable function can be defined as:

Definition 1

Let I be an interval in R. A differentiable function $f: I\rightarrow R$ is uniformly differentiable, if for any $\epsilon >0$, there is a $\delta >0$ such that for any $x,y\in I$ satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| \frac{f(x)-f(y)}{x-y}-f'(x)\right| <\epsilon \end{aligned}$$

(1)

and

$$\begin{aligned} \left| \frac{f(x)-f(y)}{x-y}-f'(y)\right| <\epsilon \end{aligned}$$

(2)

The collection of all uniformly differentiable functions on I will be denoted by UD(I). The class of uniformly differentiable function has connection with class of uniformly continuous functions which are well-known class of functions in classical analysis. The uniform continuous defined by Apostol [2] as:

Definition 2

A function $f:I\rightarrow R$ is uniformly continuous function on interval I, if for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} |{f(x)-f(y)}|<\epsilon \end{aligned}$$

(3)

Definition 3

The Caputo fractional derivative of order $\alpha$ defined by Caputo [3] as

$$\begin{aligned} {}^C{D_{}^{\alpha }}(f(t))=\frac{1}{\varGamma (n-\alpha )}\int \limits _{0}^{t}\frac{f^{(n)}(\tau )}{(t-\tau )^{n-\alpha -1}}\;d\tau \; (n-1<\alpha <n) \end{aligned}$$

(4)

The following theorem is given by Diethelm [4].

Theorem 4

Let $0<\alpha \le 1$, $a<b$ and $f\in C[a,b]$ be such that ${}^C{D_{}^{\alpha }}(f)\in C[a,b]$. Then there exist $\xi \in (a,b)$ such that

$$\begin{aligned} \frac{f(b)-f(a)}{(b-a)^{\alpha }}=\frac{1}{\varGamma (\alpha )}{}^C{D_{}^{\alpha }}(f(\xi )) \end{aligned}$$

(5)

Also some properties of Local fractional calculus was studied by Yang [5] and Yang and Gao [6]. Kachhia and Prajapati [7] introduced concept of functions of bounded fractional differential variation using the Caputo-type fractional derivative.

Definition 5

A Caputo fractional differentiable function f is absolutely fractional differentiable function on interval I, if for any $\epsilon >0$, there is a $\delta >0$ such that for an collection of pairwise disjoint intervals $\{(a_i,b_i)\}$ in I satisfying $\sum \nolimits _{i=1}^n(b_i-a_i)<\delta$,

$$\begin{aligned} \sum \limits _{i=1}^n\left| \varGamma (\alpha )\left( \frac{f(b_i)-f(a_i)}{(b_i-a_i)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(a_i))\right| <\epsilon \end{aligned}$$

(6)

and

$$\begin{aligned} \sum \limits _{i=1}^n\left| \varGamma (\alpha )\left( \frac{f(b_i)-f(a_i)}{(b_i-a_i)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(b_i))\right| <\epsilon \end{aligned}$$

(7)

where $0<\alpha \le 1$.

The Hölder continuous function defined by Gilberg and Trudinger [8] as:

Definition 6

A function $f:{R}\rightarrow C$ is said to be Hölder continuous if for all $x,y\in R$, there are non-negative real constants $M,\alpha$ such that

$$\begin{aligned} |f(x)-f(y)|\le M|x-y|^\alpha \end{aligned}$$

2 Uniformly Fractional Differentiable Functions

In this section, authors introduced the new concept of uniformly factional differentiable functions as:

Definition 7

Let I be an interval in R. A Caputo fractional differentiable function f is uniformly fractional differentiable function on I, if for any $\epsilon >0$, there is a $\delta >0$ such that for any $x,y\in I$ satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <\epsilon \end{aligned}$$

(8)

and

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))\right| <\epsilon \end{aligned}$$

(9)

where $0<\alpha \le 1$.

If we take $\alpha =1$, then Eqs. (8) and (9) reduces to Eqs. (1) and (2) respectively. The collection of all uniformly fractional differentiable functions on I will be denoted by UFD(I).

Theorem 8

A function f is uniformly fractional differentiable function on an interval I if and only if ${}^C{D_{}^{\alpha }}(f)$ is uniformly continuous on I.

Proof

Let $f:I\rightarrow R$ be uniformly fractional differentiable. Then for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^{\alpha }}\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <{\frac{\epsilon }{2}} \end{aligned}$$

(10)

and

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^{\alpha }}\right) -{}^C{D_{}^{\alpha }}(f(y))\right| <{\frac{\epsilon }{2}} \end{aligned}$$

(11)

Now for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned}&\left| {{}^C{D_{}^{\alpha }}(f(x))-{}^C{D_{}^{\alpha }}(f(y))}\right| = \nonumber \\&\left| {{}^C{D_{}^{\alpha }}(f(x))-\varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) +\varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))}\right| \end{aligned}$$

(12)

We get

$$\begin{aligned} \begin{aligned} \left| {{}^C{D_{}^{\alpha }}(f(x))-{}^C{D_{}^{\alpha }}(f(y))}\right|&\le \left| {}^C{D_{}^{\alpha }}(f(x))-\varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) \right| \\&\quad +\,\left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))\right| \end{aligned} \end{aligned}$$

(13)

By using Eqs. (10) and (11), we obtain

$$\begin{aligned} |{{}^C{D_{}^{\alpha }}(f(b_i))-{}^C{D_{}^{\alpha }}(f(a_i))}|<{\frac{\epsilon }{2}}+{\frac{\epsilon }{2}}=\epsilon \end{aligned}$$

(14)

Hence ${}^C{D_{}^{\alpha }}(f)$ is a uniformly continuous on I.

Conversely suppose that ${}^C{D_{}^{\alpha }}(f)$ is uniformly continuous on I. Let $\epsilon >0$ be given. Then there exist a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| {{}^C{D_{}^{\alpha }}(f(x))-{}^C{D_{}^{\alpha }}(f(y))}\right| <\epsilon \end{aligned}$$

(15)

Then from Theorem 4, there exist $c\in (y,x)$ such that

$$\begin{aligned} f(x)-f(y)=\frac{{}^C{D_{}^{\alpha }}(f(x))(x-y)^\alpha }{\varGamma (\alpha )} \end{aligned}$$

(16)

Since $|c-y|<\delta$, for any $\epsilon >0$, there exist a $\delta >0$ such that for any x, y in I

$$\begin{aligned} |{{}^C{D_{}^{\alpha }}(f(c))-{}^C{D_{}^{\alpha }}(f(y))}|<\epsilon \end{aligned}$$

(17)

By using Eq. (16)

$$\begin{aligned} \sum \limits _{i=1}^n\left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))\right| <\epsilon \end{aligned}$$

(18)

Again $|x-c|<\delta$, then for any $\epsilon >0$, there exist a $\delta >0$ such that for any x, y in I

$$\begin{aligned} \sum \limits _{i=1}^n\left| {{}^C{D_{}^{\alpha }}(f(x))-{}^C{D_{}^{\alpha }}(f(c))}\right| <\epsilon \end{aligned}$$

(19)

By using Eq. (16)

$$\begin{aligned} \sum \limits _{i=1}^n\left| {}^C{D_{}^{\alpha }}(f(x))-\varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) \right| <\epsilon \end{aligned}$$

(20)

Therefore f is uniformly fractional differentiable on I.

Example 9

The $\frac{1}{2}$ order Caputo derivative of function $f(t)=t$ is $2\sqrt{\frac{t}{\pi }}$ which is uniformly continuous on [0, c]. Then by Theorem 8 uniformly fractional differentiable functions on [0, c] of order $\frac{1}{2}$.

In fact, using Theorem 8, several examples of uniformly fractional differentiable functions can be constructed.

The following is motivated by the principle that differentiability implies continuity.

Theorem 10

If f is uniformly fractional differentiable function on an interval I, then f is uniformly continuous on I.

Proof

Since a function $f: I\rightarrow R$ is uniformly fractional differentiable, then if for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <\epsilon \end{aligned}$$

(21)

and

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))\right| <\epsilon \end{aligned}$$

(22)

Since ${}^C{D_{}^{\alpha }}(f)$ is bounded on I, so there exit $M>0$ such that

$$\begin{aligned} |{}^C{D_{}^{\alpha }}(f(t))|\le M\;(\forall \;t\in I ) \end{aligned}$$

(23)

Take $\delta _0=\min \{(\delta )^{\frac{1}{\alpha }},(\frac{\epsilon }{\epsilon +M})^{\frac{1}{\alpha }}\}$. Let $x,y\in I$ satisfying $|x-y|<\delta _0$.

Now

$$\begin{aligned}&\quad |f(x)-f(y)|\le \nonumber \\&\left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) (x-y)^\alpha -{}^C{D_{}^{\alpha }}(f(x))(x-y)^\alpha +{}^C{D_{}^{\alpha }}(f(x))(x-y)^\alpha \right| \end{aligned}$$

(24)

Therefore

$$\begin{aligned} \begin{aligned}&\quad |f(x)-f(y)|\le \\&\left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| {|x-y|^\alpha }+|{}^C{D_{}^{\alpha }}(f(x))|{|x-y|^\alpha } \end{aligned} \end{aligned}$$

(25)

Finally

$$\begin{aligned} |f(x)-f(y)|<\delta _0 \epsilon +M\delta _0=\delta _0(\epsilon +M)<\epsilon \end{aligned}$$

(26)

Hence f is an uniformly continuous on I.

Theorem 11

Every absolutely fractional differentiable function on I is uniformly fractional differentiable on I.

Proof

Since $f:I\rightarrow R$ is an absolutely fractional differentiable. Then for any $\epsilon >0$, there is a $\delta >0$ such that for any finite collection of pairwise disjoint intervals $\{(a_i,b_i)\}$ in I satisfying $\sum \nolimits _{i=1}^n(b_i-a_i)<\delta$,

$$\begin{aligned} \sum \limits _{i=1}^n\left| \varGamma (\alpha )\left( \frac{f(b_i)-f(a_i)}{(b_i-a_i)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(a_i))\right| <\epsilon \end{aligned}$$

(27)

and

$$\begin{aligned} \sum \limits _{i=1}^n\left| \varGamma (\alpha )\left( \frac{f(b_i)-f(a_i)}{(b_i-a_i)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(b_i))\right| <\epsilon \end{aligned}$$

(28)

In particular

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <\epsilon \end{aligned}$$

(29)

and

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))\right| <\epsilon \end{aligned}$$

(30)

Hence f is uniformly fractional differentiable function on I.

Proposition 12

If f is uniformly fractional differential function on I and if ${}^C{D^{\alpha }_{a}}(f)$ is bounded on I, then f is Hölder continuous on I.

Proof

Let f is uniformly fractional differential function. Then for x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <\epsilon \end{aligned}$$

(31)

and

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))\right| <\epsilon \end{aligned}$$

(32)

Since ${}^C{D_{}^{\alpha }}(f)$ is bounded on I, so there exit $M>0$ such that

$$\begin{aligned} |{}^C{D_{}^{\alpha }}(f(t))|\le M\;(\forall \;t\in I ) \end{aligned}$$

(33)

Now

$$\begin{aligned} |f(x)-f(y)|= & {} \left| \left( \frac{f(x)-f(y)}{(x-y)^{\alpha }}\right) (x-y)^{\alpha }-{}^C{D^{\alpha }_a}(f(y))(x-y)^{\alpha }+{}^C{D^{\alpha }_a}(f(y))(x-y)^{\alpha }\right| \\\le & {} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^{\alpha }}\right) -{}^C{D^{\alpha }_a}(f(y))\right| |x-y|^{\alpha }+|{}^C{D^{\alpha }_a}(f(y))||x-y|^{\alpha }\\\le & {} (\epsilon +M)|x-y|^{\alpha } \end{aligned}$$

Hence f is Hölder continuous function on R.

Theorem 13

The space UFD(I) of uniformly fractional differentiable functions on interval I is a vector space with pointwise operations.

Proof

Let $f,g\in UFD(I)$. Then for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <\frac{\epsilon }{2} \end{aligned}$$

(34)

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <\frac{\epsilon }{2} \end{aligned}$$

(35)

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{g(x)-g(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(g(x))\right| <\frac{\epsilon }{2} \end{aligned}$$

(36)

and

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{g(x)-g(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(g(x))\right| <\frac{\epsilon }{2} \end{aligned}$$

(37)

Now for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned}&\left| \varGamma (\alpha )\left( \frac{(f+g)(x)-(f+g)(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((f+g)(x))\right| =\nonumber \\&\qquad \left| \varGamma (\alpha )\left( \frac{(f(x)+g(x))-(f(y)+g(y))}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((f(x)+(g(x))\right| \end{aligned}$$

(38)

Then

$$\begin{aligned}&\quad \left| \varGamma (\alpha )\left( \frac{(f+g)(x)-(f+g)(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((f+g)(x))\right| \le \nonumber \\&\left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| +\left| \varGamma (\alpha )\left( \frac{g(x)-g(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(g(x))\right| \end{aligned}$$

(39)

By using Eqs. (34) and (35) the Eq. (39) reduces to

$$\begin{aligned} \begin{aligned} \sum \limits _{i=1}^n\left| \varGamma (\alpha )\left( \frac{(f+g)(x)-(f+g)(x)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((f+g)(x))\right| <\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon \end{aligned} \end{aligned}$$

(40)

Similarly by using Eqs. (36) and (37) we obtain

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{(f+g)(x)-(f+g)(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((f+g)(y))\right| <\epsilon \end{aligned}$$

(41)

Hence $f+g\in UFD(I)$. Now let $f\in UFD(I)$ and $k\in C$. Then for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(x))\right| <\frac{\epsilon }{k}, \end{aligned}$$

(42)

and

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}(f(y))\right| <\frac{\epsilon }{k},\end{aligned}$$

(43)

Now for any $\epsilon >0$, there is a $\delta >0$ such that for any x, y in I satisfying $|x-y|<\delta$,

$$\begin{aligned} \begin{aligned} \left| \varGamma (\alpha )\left( \frac{(kf)(x)-(kf)(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((kf)(x))\right| =\\ \left| k\varGamma (\alpha )\left( \frac{f(x)-f(y)}{(x-y)^\alpha }\right) -k\;{}^C{D_{}^{\alpha }}((f(x))\right| \end{aligned} \end{aligned}$$

(44)

By using Eq. (42) the above equation reduces to

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{(kf)(x)-(kf)(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((kf)(x))\right| <k\frac{\epsilon }{k}=\epsilon \end{aligned}$$

(45)

Similarly by using Eq. (43) we obtain

$$\begin{aligned} \left| \varGamma (\alpha )\left( \frac{(kf)(x)-(kf)(y)}{(x-y)^\alpha }\right) -{}^C{D_{}^{\alpha }}((kf)(y))\right| <\epsilon \end{aligned}$$

(46)

Thus $kf\in UFD(I)$.

Therefore the space UFD(I) of uniformly fractional differentiable functions on I is a vector space with pointwise operations.