# To prove the existence of a constant $c$ in a function

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 $x$

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

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.
• 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? – Jason May 8 '17 at 16:14
• @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. – Michael May 8 '17 at 21:17