Find all functions $f:\mathbb{R}\to\mathbb{R}$ such as $x\left[f(x+y)-f(x-y)\right]=4yf(x)$

Question

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such as $$x\left[f(x+y)-f(x-y)\right]=4yf(x)$$

My attempt:

Let $x=0\Rightarrow 4yf(0)=0\Rightarrow f(0)=0.$

Let $y=x\Rightarrow xf(2x)=4xf(x)\Rightarrow f(2x) = 4f(x).$

Then I don't know what should I do next :((

Thanks a lot.

Answer 1

Call function $f: \mathbb{R}\to \mathbb{R}$ good if $\forall \,x,y\in \mathbb{R}$ satisfies $$x\big(f(x+y)-f(x-y)\big) = 4yf(x),$$ and let $F= \{f: \mathbb{R}\to \mathbb{R}; f \;{\rm is\;good}\}$. Then $F$ is a vector space in a set of all real function with real variable. Clearly $f(x) = x^2$ is in $F$ so $\dim F\geq 1$ and let us prove $\dim F =1$ which leads to each good function is of form $cx^2$.

Notice that map $M: F\to \mathbb{R}$ defined with $f\mapsto f(1)$ is linear. If we prove $M$ is injective, then $$\dim F = \dim ({\rm Im} (M)) = \dim( \mathbb{R})=1$$ and we are done.

So let $f\in \ker (M)$, then we have $f(1)=0$ and as you already proved we have $f(0)=0$ and $f(2x)=4f(x)$.

• Claim 1: For all $x$ we have $f(x)= f(2-x)$ and $f(1-x)=f(x)$:

Let $x=1$, then we get $$f(1+y)=f(1-y)\implies f(t)= f(2-t)$$ where $t=1+y$, so $t$ can be any real number, and now we have $$4f(x) =f(2x) = f(2-2x) = 4f(1-x)\implies f(1-x)=f(x)$$

• Claim 2: For all $x$ we have $f(x+1) = f(x)$ and $f(x-1) = f(3-x)$:

$$ f(1+x) = f(2-(1+x) ) = f(1-x) = f(x)$$ and $$ f(x-1) = f(2-(x-1)) = f(3-x)$$

• Claim 3: For all $a$ we have $f(3a)= 9f(a)$:

Let $x=2a$ and $y=a$: $$2a\big(f(3a)-f(a)\big)= 4af(2a)= 16af(a)\implies f(3a)-f(a)= 8f(a)$$ and thus the result.

• Finally:

Now let $y=1$, then we have $$x\big(f(x+1)-f(x-1)\big) = 4f(x)$$ By claim 2 we have $$x(f(x)) - f(3-x)) = 4f(x)$$ Let $x=3t$, then we have $$3t \big(f(3t) -f(3(1-t))\big) = 4f(3t) $$ then by claim 3 we have $$27t \big(f(t) -f(1-t)\big) = 36f(t) $$ But LHS is by claim 1 equal 0 and thus $f(t)=0$ for all $t$. So $\ker (M)= \{0\}$ and thus a conclusion.