# Problem

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 $$f(x)$$.

# Attempt

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?

• 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. Jun 1 '19 at 5:17
• You need to be clear. For one function $f$ your equation holds for one value $a$? Or for all values $a\ne 0$? Jun 1 '19 at 12:17
• Sep 13 '20 at 6:36

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 $2$

• 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. Jun 1 '19 at 8:44
• @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! Jun 1 '19 at 8:53
• @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. Jun 1 '19 at 9:23
• Kavi.The problem is I trust your(!) judgment. Greetings. Jun 1 '19 at 10:43