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## Understand the problem

Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function such that$\frac{1}{2y}\int_{x-y}^{x+y}f(t)\, dt=f(x)\qquad\forall~x\in\mathbb{R}~\&~y>0$Show that there exist $a,b\in\mathbb{R}$ such that $f(x)=ax+b$ for all $x\in\mathbb{R}$.
##### Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 4

### Calculus

8.5 out of 10

##### Suggested Book

Do you really need a hint? Try it first!

Do you know Taylor series expansion good then see….. $f(x+h) = f(x) + f'(x)h+f”(x)\frac{h^2}{2} + f”'(x)\frac{h^3}{3!}+\cdots$ $f(x-h) = f(x) – f'(x)h+f”(x)\frac{h^2}{2} – f”'(x)\frac{h^3}{3!}+\cdots$

adding previous expression we get  $\frac{f(x+h) – 2f(x) + f(x-h)}{h^2} = f”(x) + 2\frac{f””(x)}{4!}h^2+\cdots$  taking the limit of the above equation as h goes to zero gives  $\Rightarrow f”(x) = \lim_{h\to0} \frac{f(x+h) – 2f(x) + f(x-h)}{h^2} \,.$

Since $f$ is continuous, there exists $F$ such that $F' = f.$ The given identity becomes  $F(x+y)-F(x-y) = 2yf(x).$ Fix $x \in \mathbb{R}$ and differentiate the above identity with respect to $y$ and get $f(x+y)+f(x-y)=2f(x).$

now we already get ,  If $f$ is twice differentiable at $x$ then $\lim_{h\to 0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x).$ by rearranging our given problem we get  that $f''(x)=0$ Therefore , $f(x)=ax+b$ for some $a,b\in\mathbb{R}.$

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