A function such that $f(x) = \lim_{t\to0}\frac{1}{2t}\int_{x-t}^{x+t} sf'(s)\,ds$ for all $x$

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?

• This being your 19th question, you should know better than post blurry screenshots of problems. See math notation guide. – user147263 Jan 21 '15 at 15:58
• @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. – 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 $$x\,f'(x)=f(x),\quad f(\sqrt2)=2.$$ $$\frac{f'}{f}=\frac{1}{x}\implies (\log f)'=\log x+c\implies f(x)=C\,x.$$