Let $f:\mathbb R\to\mathbb R$ be a function with continuous derivative such that $f(\sqrt{2})=2$ and $$f(x) = \lim_{t\to0}\frac{1}{2t}\int_{x-t}^{x+t} sf'(s)\,ds$$ for all $x\in\mathbb R$. Find $f(3)$.

I guess Fundamental theorem of Calculus needs to be used to solve this.

Taking derivative of x on both sides I simplified the integral to

$(x+t)f'(x+t) - (x-t)f'(x-t) $

The equation becomes:

$f'(x) = \lim (1/2t)(x+t)f'(x+t) - (x-t)f'(x-t) $ as t tends to 0.

This is leading me nowhere. Any ideas on how to tackle this problem?

  • $\begingroup$ This being your 19th question, you should know better than post blurry screenshots of problems. See math notation guide. $\endgroup$ – user147263 Jan 21 '15 at 15:58
  • 1
    $\begingroup$ @Fundamental Thanks for editing my post. Yes, this is my 19th question, all within a span of about a month! That is because I have an exam coming up and I could really use some help. But right now I really don't have the time to sit back and learn MathJax. I'll definitely learn it once I am done with my exam. $\endgroup$ – Deepabali Roy Jan 21 '15 at 16:07

Lemma. If $g$ is a continuous function, then $$ \lim_{h\to0}\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds=g(x). $$ Proof. We may assume $h>0$. $$ \Bigl|g(x)-\frac{1}{2\,h}\int_{x-h}^{x+h}g(s)\,ds\Bigr|=\frac{1}{2\,h}\Bigl|\int_{x-h}^{x+h}(g(x)-g(s))\,ds\Bigr|\le\frac{1}{2\,h}\int_{x-h}^{x+h}|g(x)-g(s)|\,ds. $$ Use that $g$ is continuous at $x$ to show that the last expression converges to $0$ as $h\to0$.

Let's return to the original question. Since $x\,f'(x)$ is continuous, we have $$ \lim_{t\to0}\frac{1}{2\,t}\int_{x-t}^{x+t} s\,f'(s)\,ds=x\,f'(x). $$ (You can get the same result integrating by parts.) All is left is to solve the ODE $$ x\,f'(x)=f(x),\quad f(\sqrt2)=2. $$ $$ \frac{f'}{f}=\frac{1}{x}\implies (\log f)'=\log x+c\implies f(x)=C\,x. $$

  • $\begingroup$ I didn't understand how you simplified the expression to xf′(x). I am not aware of the theorem used here. Could you please explain this a little, or share a link I could read to understand this? Thanks a lot! $\endgroup$ – Deepabali Roy Jan 21 '15 at 16:25
  • $\begingroup$ See the edited answer. $\endgroup$ – Julián Aguirre Jan 21 '15 at 16:33
  • $\begingroup$ I can't see any edits to the answer :/ $\endgroup$ – Deepabali Roy Jan 21 '15 at 16:35
  • $\begingroup$ Look now. I did not press "Save edits". $\endgroup$ – Julián Aguirre Jan 21 '15 at 16:38
  • $\begingroup$ The problem seemed so difficult to solve. Thank you :) $\endgroup$ – Deepabali Roy Jan 21 '15 at 16:43

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