Let $f:\Bbb{R}\to \Bbb{R} $ a continuous function s.t. $f(x+y)+f (x-y)=2f(x)+2f(y) $. Find $f$.

Question

Let $f:\Bbb{R}\to \Bbb{R} $ a continuous function s.t. $f(x+y)+f (x-y)=2f(x)+2f(y) $. Find $f$.

My idea: Let $x=y$. Then $f (2x)+f(0)=4f(x)$. I tried to use this recurrence but it doesn't work.

Answer 1

As already noted in the comments, $f(0)=0$ (take $x=y=0$), and $f(-z)=f(z)$ (take $x=0$, $y=z$). Further, $f(az)=a^2 f(z)$ for natural $a$ (it is proven by induction on $a$ taking $x=az$, $y=z$), and therefore for any rational $a$ (because $f(z)=f(az)/a^2$). By continuity, it holds for any real $a$.

So $f(x)=x^2 f(1)$.

Answer 2

As I'm new I can't comment but here is something you should consider:

Let $x=0\ ,\ f(y)+f(-y)=2f(y)\ , so\ f(-y)=f(y)$

Also I proved for integers:

Let $x=k-1\ ,y=1 : f(k)+f(k-2)=2f(k-1)+2f(1)$

Now let $k=0 : f(0)+f(2)=2f(1)+2f(1)\Leftrightarrow f(2)=4f(1)$

If for every integer $0\le l \le k$ we have $f(l)=l^2f(1)$ then:

$f(k+1)+f(k-1)=2f(k)+2f(1)\Leftrightarrow f(k+1)+(k-1)^2f(1)=2k^2f(1)+2f(1)\Leftrightarrow f(k+1)=f(1)(2k^2+2-(k-1)^2)\\ \qquad \qquad \; \ \ =f(1)(2k^2+2 -(k^2-2k+1))\\ \qquad \qquad \; \ \ =f(1)(k^2+2k+1)\\ \qquad \qquad \; \ \ =(k+1)^2f(1)$

Therefore by induction we get $f(x)=x^2f(1)$