Let $f$ be a Riemann integrable function defined on $[-2,2]$. Define a function $F \colon (-1,1) \to \mathbb{R}$ by $$F(h)=\int_0^1 h | f(x+h)-f(x)|\, dx.$$ Show that the derivative $F'(0)$ exists.

I started form $\lim_{h\to 0}$ $\frac{F(h)-F(0)}{h}$

By definition, $F(0)=0$ since it $F(0)$ becomes a definite integral of zero.

$\lim_{h\to 0}$ $\frac{F(h)-F(0)}{h} = \lim_{h\to 0}$ $\frac{F(h)}{h} = $$\lim_{h\to 0}\int_0^1 |f(x+h)-f(x)| dx$. Them I have no clue to continue, can anyone give me some hints?

  • $\begingroup$ Do you know about the $L^p$ spaces? $\endgroup$ – user258700 Apr 30 '16 at 10:44
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    $\begingroup$ Hint: If $f$ were increasing, then could you solve it (first step: can you write this as two integrals over smaller domains)? $\endgroup$ – Michael Burr Apr 30 '16 at 10:50
  • $\begingroup$ A possible idea is to approximate $f$ with a (uniformly) continuous function: math.stackexchange.com/questions/1139040/… $\endgroup$ – Siminore Apr 30 '16 at 11:03
  • $\begingroup$ @AhmedHussein I have seen $L^p$ in my functional analysis, but we use it as an example illustrate the convergence of cauchy sequence. $\endgroup$ – Belive Apr 30 '16 at 11:05
  • $\begingroup$ @MichaelBurr Like $\lim_{h\to 0}\int_0^1 |f(x+h)-f(x)| dx$ =$\lim_{h\to 0}\int_0^c f(x+h)-f(x) dx+\int_c^1 f(x+h)-f(x) dx$ ? $\endgroup$ – Belive Apr 30 '16 at 11:08

Sine $f$ is integrable, given $\epsilon>0$ there is a partition $P=\{x_0,x_1,\dots,x_n\}$ of $[0,1]$ such that $0\le\int_0^1f-L(P)\le\epsilon$, where $L(P)$ is the lower sum of $f$ associated to the partition $P$. Define $g\colon[0,1]\to\mathbb{R}$ as $$ g(x)=\inf_{x_{i-1}\le x<x_i}f(x)\quad\text{on}\quad[x_{i-1},x_i),\quad 1\le i\le N,\quad g(x_N)=g(x_{N-1}). $$ Then $g$ is integrable, $\int_0^1g=L(P)$ and $$ 0\le f(x)-g(x),\quad0\le\int_0^1(f-g)\le\epsilon. $$ Then $$\begin{align} \int_0^1|f(x+h)-h(x)|\,dx&\le\int_0^1(|f(x+h)-g(x+h)|+|g(x+h)-g(x)|+|g(x)-f(x)|)\,dx\\ &=\int_0^1((f(x+h)-g(x+h))+|g(x+h)-g(x)|+(f(x)-g(x)))\,dx\\ &\le2\,\epsilon+\int_0^1|g(x+h)-g(x)|\,dx. \end{align}$$ All is left is to show that the last integral converges to $0$ as $h\to0$. This is easy, because $g$ is piecewise constant.

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  • $\begingroup$ Thank you for your answer. Do we have definition for $g(x+h)$ when$x\lt 1-h$?, since we need g(x+h) to be well defined for $x\lt 1-h$ so as to define $\int_0^1 (f-g)$ ? $\endgroup$ – Belive Apr 30 '16 at 11:58
  • $\begingroup$ The last integral is $\le 2h*2M*N$, where $M=\sup|g(x)|$? Since$|g(x+h)-g(x)|=0 $ on $[x_i-1+h,x_i-h] $ Thus it coverges to zero. $\endgroup$ – Belive Apr 30 '16 at 12:11
  • $\begingroup$ Define it to be equal to the value in $x_N=1$. $\endgroup$ – Julián Aguirre Apr 30 '16 at 12:41

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