Let $f(x)$ be a differentiable function with $f(0)=1$ and $f(1)=2$. For any $a,x$ where $a \neq 0$, it holds $$\frac{1}{2a}\int_{x-a}^{x+a}f(t)dt=f(x).$$ Find $f(x)$.


Since $$f(x)=\frac{\displaystyle\int_{x-a}^{x+a}f(t)dt}{2a}$$ holds for all $x \in \mathbb{R}$ and $a \neq 0$. One can fix $x$ and take the limit as $a \to 0$. Thus $$\lim_{a \to 0}f(x)=\lim_{a \to 0}\frac{\displaystyle\int_{x-a}^{x+a}f(t)dt}{2a}=\frac{f(x+a)+f(x-a)}{2},$$ where we applied L'Hopital's rule. Thus $$f(x)=\frac{f(x+a)+f(x-a)}{2}.$$ How to go on from here?

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    $\begingroup$ In the line where you calculate the limit, you cannot end up with $a$ remaining on the r.h.s. What you argument shows is that $f(x)=f(x)$. Try differentiating both sides of the equation you are given. $\endgroup$ – uniquesolution Jun 1 at 5:17
  • $\begingroup$ You need to be clear. For one function $f$ your equation holds for one value $a$? Or for all values $a\ne 0$? $\endgroup$ – GEdgar Jun 1 at 12:17

The answer is that $f(x)=1+x$.

By differentiating both sides you get $$\frac{f(x+a)-f(x-a)}{2a}=f'(x)$$ and this holds for every $a$, so differentiating by $a$: $$(f'(x+a)+f'(x-a))2a-2(f(x+a)-f(x-a))=0$$ Dividing by $a$ you get: $$f'(x+a)+f'(x-a)=\frac{f(x+a)-f(x-a)}{a}=2f'(x)$$ Differentiating by $a$ again gives $$f''(x+a)-f''(x-a)=0$$ and since $x,a$ are arbitrary, this implies that $f''(x)$ is constant. Therefore, $f(x)=\alpha+\beta x+\gamma x^2$ for some $\alpha,\beta,\gamma$. Invoking the initial data we deduce that $f(x)=1+\beta x+(1-\beta)x^2$ for some $\beta$. Integrating around zero, we get $$1=f(0)=\frac{1}{2a}\int_{-a}^a(1+\beta t+(1-\beta)t^2)\,dt=1-\frac{a^2}{3}(\beta-1)$$ which implies that $\beta=1$, and so the function $f(x)$ must be equal to $1+x$.


A bit late this answer but I think it is worth mentioning it:

Applying differentiation with respect to $a$ to

  • $\int_{x-a}^{x+a}f(t)dt=2af(x)$ gives $$2f(x) = f(x+a) + f(x-a)$$

$f$ is differentiable, so differentiating wrt. $a$ again gives $$0 = f'(x+a) - f'(x-a)$$

Since this holds true for all $a,x \in \mathbb{R}$, it is also true for $a = x$: $$f'(2x) = f'(0) \Rightarrow f' = const. \stackrel{f(0)=1,f(1)=2}{\Longrightarrow} f(x) = 1+x$$


Differentiate to get $f'(x)=\frac {f(x+a)-f(x-a)} {2a}$ for all $x$ and $a$. Conclude from this that $f(b)-f(a)=(b-a)f'(\frac {a+b} 2)$ for all $a$ and $b$. Use the following to complete the solution: If $f(b)-f(a)=(b-a)f'(\frac{a+b}{2})$ prove that any such function is a polynomial of degree $2$

  • $\begingroup$ Kavi.A question, please.f is differentiable once, ok. But in the link and proof above f is differentiated twice!! with respect to a. Your thoughts?Thank you. $\endgroup$ – Peter Szilas Jun 1 at 8:44
  • $\begingroup$ @PeterSzilas In the third line of the answer by uniquesolution the left is differentiable. Hence the right side (which is $f'(x)$) is also differentiable which means $f$ is twice differentiable. In fact the function has derivatives of all orders! $\endgroup$ – Kabo Murphy Jun 1 at 8:53
  • $\begingroup$ @Kavi.Thanks.3 rd line LHS is differentiable, ok, so is RHS.Then comes: f'(x+a)+f'(x-a)=(f(x+a)-f(x-a))/a=2f'(x). The middle of this equality is differentiable, hence the LHS and RHS are also.But: Can one infer since [f:(x+a)+f'(x-a)] , i.e. the sum is differentiable, each!! term in the bracket is differentiable, to arrive at f''(x+a)-f''(x-a)=0?Thank you, again. $\endgroup$ – Peter Szilas Jun 1 at 9:23
  • $\begingroup$ Kavi.The problem is I trust your(!) judgment. Greetings. $\endgroup$ – Peter Szilas Jun 1 at 10:43

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