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Let $f: R \to R$ be a continuous function.Suppose $$ f(x) = \frac{1}{t}\int_{0}^{t} (f(x+y)-f(y))dy$$ for all $x \in R $ and $t>0$. Then show that there exists a constant $c$ such that $f(x)=cx$ for all $x$

At first I want to ask how shall I integrate the function give above? And also how am I supposed to prove the relationship of $f(x)$ with constant $c$ for all value of $x$

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  • $\begingroup$ Do you know Newton-Leibinitz's Rule? $\endgroup$ – Jaideep Khare May 8 '17 at 15:39
  • $\begingroup$ Do you mean Leibinitz's Rule for integration? @JaideepKhare $\endgroup$ – Iti Shree May 8 '17 at 15:40
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    $\begingroup$ Or you can multiply both sides by $t$ and differentiate with respect to $t$. $\endgroup$ – Michael May 8 '17 at 15:44
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    $\begingroup$ Okay thanks I am gonna try it now. @Michael $\endgroup$ – Iti Shree May 8 '17 at 15:45
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    $\begingroup$ @Omnomnomnom Why? The LHS is $tf(x)$ and the RHS is $\int_0^tg(y)dy$ for some continuous function $g$. $\endgroup$ – Jason May 8 '17 at 15:49
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Hint: $$ f(x) = \frac{d}{dt}\int_{0}^{t}f(x+y) - f(y)dy = f(x+t) - f(t) $$ $$ \frac{d}{dx}f(x) = \frac{1}{t}\frac{d}{dx}\int_{0}^{t}f(x+y) - f(y)dy = \frac{1}{t}(f(x+t) - f(x)) = f(t)/t $$ Therefore the derivative is constant.

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    $\begingroup$ I'm not quite sure how you got the second equality in your second line, care to explain? Also, how do you know for sure that $f$ is differentiable? $\endgroup$ – Jason May 8 '17 at 16:14
  • $\begingroup$ A simple way to earn more respect is to reference related work. This also holds when writing research papers. It certainly can't hurt to write something like "Following the suggestion of Michael..." or "similar to the suggestion of..." , and it even promotes a positive and collegial atmosphere. $\endgroup$ – Michael May 8 '17 at 16:22
  • $\begingroup$ @Jason : What Marc is not telling you is that, assuming we can bring the derivative inside the integral and differentiate $f$, then $$ \frac{d}{dx} \int_0^t [f(x+y)-f(y)]dy = \int_0^t f'(x+y) dy = f(x+t)-f(x)$$ which also assumes we can integrate the derivative. Without making all these assumptions, it is better to just work directly with the first equality and look at rational numbers. $\endgroup$ – Michael May 8 '17 at 21:17
  • $\begingroup$ @Michael Thank you, I realized this soon after my question but since there were missing steps I decided to leave it up (plus, as you also have pointed out, there still is the issue of not justifying differentiability). $\endgroup$ – Jason May 8 '17 at 21:20
  • $\begingroup$ Thank you so much for your help Marc, had really hard time understanding the question. $\endgroup$ – Iti Shree May 10 '17 at 17:14

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