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Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. $\forall_ {\delta>0}$ we will denote: $$F_\delta(x)=\frac{1}{2\delta}\int_{x-\delta}^{x+\delta}f(t)dt$$

Prove that $\forall x\in\mathbb{R}\Bigg| \lim_{\delta\rightarrow 0} F_\delta (x)=f(x)$.
My attempt:
I don't have an idea of solving this other than $F_\delta (x)=\frac{1}{2\delta}( F(x+\delta)-F(x-\delta))$, where F is the anti-derivative of f. But now the only thing I can think of is that it looks like mean value, for some reason. Other than that, couldn't think of anything.

A field
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  • A proof of Lebesgue’s differentiation theorem – Matcha Latte Jul 09 '23 at 07:54
  • There are a lot of things there that I have never seen in my life. And I dont think the 2 questions are even similar. – A field Jul 09 '23 at 07:56
  • Lebesgue Differentiation theorem – Matcha Latte Jul 09 '23 at 07:56
  • $(x-\delta,x+\delta)=B(x,\delta)\subseteq\Bbb R^n$ for $n=1$ – Matcha Latte Jul 09 '23 at 07:58
  • So their prove is for n-dimensional space? – A field Jul 09 '23 at 07:59
  • And there isn't an easier prove of this(for 1 dimensional space)? Because I can not use those things in my exam, if I wanted to. My professor says that we can use only what we learned so far. This is a question from Infi B exam(we have A,B,C). And you are using tihngs from Calc C. – A field Jul 09 '23 at 08:04
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    Rewrite your formula for $F_{\delta}(x)$ as $\frac{1}{2} (]\frac{F(x+\delta) - F(x)}{\delta}] + [\frac{F(x) - F(x - \delta) }{\delta}])$ and note that as $\delta$ goes to $0$ each of the terms in brackets goes to $F'(x)$ from the limit definition of the derivative... – leslie townes Jul 09 '23 at 08:07
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    $F_\delta(x)-f(x)=\frac{1}{2\delta}\int_{x-\delta}^{x+\delta}[f(t)-f(x)]dt$. Just use continuity of $f$ at $x$ to finish. Lebsgue's differentiation theorem is surely an overkill for this. – Kavi Rama Murthy Jul 09 '23 at 08:17
  • Not only is the Lebesgue differentiation theorem overkill, it's also not good enough. It only guarantees the result almost everywhere whereas we want it to hold everywhere. – FShrike Jul 09 '23 at 10:15
  • to op: Lebesgue differentiation is very likely not in this "Calc C" course either, it comes from measure theory which is more advanced than 'calculus' – FShrike Jul 09 '23 at 10:18
  • Yes. I have to take measure theory in the second semester of my second year(TMRW, hopefully, I'll have my last exam of my first year - I will do a Calc B exam, and I'm stressed af) – A field Jul 09 '23 at 10:21
  • geetha290km's approach is the way to go. Or equivalently, since $f$ is continuous at $x$, for each $\epsilon > 0$, there is some $\delta > 0$ with $f(x) - \epsilon < f(t) < f(x) + \epsilon$ when $x-\delta < t < x + \delta$. Recall that if $u(t) < v(t)$ for all$t \in[a,b]$, then $\int_a^b u(t)dt \le \int_a^b v(t)dt$. – Paul Sinclair Jul 10 '23 at 20:28
  • @PaulSinclair Hey, could you tell me why you think people didn't upvote my question? Or how do I write questions so that people will upvote them? Im kinda new to this platform, and I'm trying to do better. – A field Jul 14 '23 at 20:25
  • No, I cannot. I'm afraid I rarely upvote questions. This is a failing on my part, but while I find it easy to decide if an anaswer is worthy of an upvote or not, questions are not so clear to me. I'll upvote this one mostly because you have now demonstrated a desire to ask good questions. You have the basics covered, though perhaps people may feel that your description of your efforts doesn't go very far, and thus indicates you hadn't put much thought into it. I'm more sanguine about that, acknowledging how hard it is to put such things in words. – Paul Sinclair Jul 15 '23 at 12:08

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