What method is used to find the expression of a function?

Question

I've found some difficulties in this exercise please could you give me help?

Let $f$ be a continuous function in $\mathbb R$ such that

$$\forall(x;y)\in\mathbb R, f (x+y) + f(x-y) = 2(f(x)+f(y)).$$

1. Calculate $f(0)$ then show that $f$ is an even function.
2. Let $x\in\mathbb R$. Prove that $\forall n\in\mathbb Z,f(nx)=n^2f(x)$.
3. Let $a=f(1)$. Prove that $\forall r\in\mathbb Q,f(r)=r^2 a$
4. Deduce the expression of $f \in \mathbb R$.

Answer 1

1. a) Take $x = y = 0$. b) Take $y=0$
2. Take $y=x$, this will give a formula for $f(2x)$. Now induct.

3. let $a = f(1)$, then $f(n) = n^2a$ for all $n \in \mathbb{N}$ by (2). Now for $r = n/m \in \mathbb{Q}$, note that $$ m^2f(r) = f(mr) \text{ by Part (1)} $$ $$ = f(n) = n^2a $$ Hence, $f(r) = r^2 a$

4. Since $f$ is continuous, for any $x\in \mathbb{R}$, choose a sequence $q_n \in \mathbb{Q}$ such that $q_n \to x$, then $$ q_n^2 a = f(q_n) \to f(x) \Rightarrow f(x) =x^2 a $$

Answer 2

To expand a bit for answer 2. If we take $x=y$ we get easily $$ f(2x)+f(0)=4f(x) $$ but since in 1. we have shown that $f(0)=0$ we get easily $f(2x)=4f(x)$. Now let's assume 2. is valid for $n$ and let's show it is true for $n+1$. Since we have shown it for $n=2$ then it is true for all $n$. Please note that this demonstration is restricted to $n>2$ but can be easily extended to negative values.

So we write our initial equation making following substitutions: $x=ns$ and $y=s$. We get easily (I will skip some easy steps) $$ f((n+1)s) +f((n-1)s)=2f(ns)+2f(s) $$ and using $f(ns)=n^2f(s)$ we get $$ f((n+1)s)+n^2f(s)+f(s)-2nf(s)=2f(s)+2n^2f(s) $$ that can be simplified to $$ f((n+1)s)=n^2f(s)+2nf(s)+f(s) = (n+1)^2f(s) $$ that is exactly what we wanted to show.