5
$\begingroup$

Can you please, check if my proof is correct?

Suppose that $f:[a,b]\to \Bbb{R}$ is continuous and $F(x)=\int^{x}_{a}f(t)dt$, then $F\in C^{1}[a,b]$ and $$\dfrac{d}{dx}\int^{x}_{a}f(t)dt:=F'(x)=f(x)$$

MY PROOF: Credits to Aweygan for the correction

Let $x_0\in[a,b]$ and $\epsilon>0$ be given. Since $f$ is continuous at $x_0$ then, there exists $\delta>0$ such that $|t-x_0|<\delta$ implies $$|f(t)-f(x_0)|<\epsilon.$$ Thus, $$f(x_0)=\dfrac{1}{x-x_0}\int^{x}_{x_0}f(x_0)dt,\;\;\text{where}\;\;x\neq x_0.$$ For any $x\in (a,b),$ with $0<|x-x_0|<\delta,$ such that $x_1=\min\{x,x_0\}$ and $x_2=\max\{x,x_0\}$. So, we have \begin{align}\left| \dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) \right|&= \left| \dfrac{1}{x-x_0}\int^{x}_{x_0}(f(t)-f(x_0))dt \right| \\&\leq \dfrac{1}{|x-x_0|}\int^{x}_{x_0} \left|f(t)-f(x_0) \right|dt\\&\leq \dfrac{1}{|x-x_0|}\int^{x_2}_{x_1} \left|f(t)-f(x_0) \right|dt\\&< \dfrac{1}{|x-x_0|}\epsilon|x_1-x_2| \\&\leq \dfrac{1}{|x-x_0|}\epsilon|x-x_0| =\epsilon \end{align} Hence, $$F\in C^{1}[a,b]\;\;\text{and}\;\;\dfrac{d}{dx}\int^{x}_{a}f(t)dt:=F'(x)=f(x)$$

$\endgroup$
11
  • $\begingroup$ It's not sufficient to prove the limit is $f(x)$ only from one side. $\endgroup$
    – egreg
    Jan 14 '19 at 13:33
  • $\begingroup$ You still only have differentiability from the right, you need to show differentiability from the left (i.e. when $-\delta<h<0$). $\endgroup$
    – Aweygan
    Jan 14 '19 at 13:47
  • $\begingroup$ You're still computing a one-sided limit. $\endgroup$
    – egreg
    Jan 14 '19 at 13:49
  • $\begingroup$ It's still not quite correct. I'm all but certain you'll need to handle the two cases $h>0$ and $h<0$ separately, or introduce some new variables to avoid this. $\endgroup$
    – Aweygan
    Jan 14 '19 at 13:55
  • $\begingroup$ @Aweygan: I want to rewrite it, now. Some moments, please! $\endgroup$ Jan 14 '19 at 13:56
2
$\begingroup$

It's essentially correct, but you should either split up the last part of the proof into the cases where $x<x_0$ and $x_0<x$, or write $x_1=\min\{x,x_0\}$, $x_2=\max\{x,x_0\}$ and do the following:

\begin{align} \left| \dfrac{F(x)-F(x_0)}{x-x_0}-f(x_0) \right|&= \left| \dfrac{1}{x-x_0}\int^{x}_{x_0}(f(t)-f(x_0))dt \right| \\ &\leq \dfrac{1}{|x-x_0|}\int^{x_2}_{x_1} \left|f(t)-f(x_0) \right|dt \\&< \dfrac{1}{|x-x_0|}\epsilon|x-x_0| \\ &=\epsilon. \end{align}

$\endgroup$
2
  • $\begingroup$ Thanks a lot for the explanation (+1) $\endgroup$ Jan 14 '19 at 15:07
  • $\begingroup$ You're welcome. Glad to help! $\endgroup$
    – Aweygan
    Jan 14 '19 at 15:07
2
$\begingroup$

You can also directly use the mean value theorem for integrals: $$ \lim_{h\downarrow0}\frac{\int_{a}^{x}f(t)dt-\int_{a}^{x+h}f(t)dt}{h}=\lim_{h\downarrow0}\frac{\int_{x}^{x+h}f(t)dt}{h}=\lim_{h\downarrow0}\frac{f(c_{h})h}{h} $$ where $c_{h}$ is between $x$ and $x+h$. Therefore, $\lim_{h\downarrow0}c_{h}=x$. By continuity, $\lim_{h\downarrow0}f(c_{h})=f(x)$.

$\endgroup$

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.