3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Do you know about the $L^p$ spaces? $\endgroup$ – user258700 Apr 30 '16 at 10:44
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.