# Help proving function is differentiable in $\mathbb{R}$

Let function $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable in $a=0$. And also let:

$$f(x+y)=f(x)(f(y))^2$$

I have to prove that the function is differentiable on $\mathbb{R}$.

I wonder if i can even use the definition by limit for this kind of task.

$$\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}$$ has to exist in proving its differentiable.

$$\lim_{h\rightarrow0}\frac{f(x)f(h)^2-f(x)}{h}=\lim_{h\rightarrow0}\frac{f(x)(f(h)^2-1)}{h}$$ That is what I get but how can I go further, where can I use that it's differentiable at $a=0$

And is this the right approach, any help would be appreciated.

• Are you sure of your functional equation? It seems to imply that $f$ is a constant. – lulu Aug 30 '17 at 17:53
• what do you mean with constant? – MathIsTheWayOfLife Aug 30 '17 at 17:55
• I mean $f(x)=c$ for some constant $c$. Note that $f(x+y)=f(y+x)$ so you get $f(x) \, \left( f(y)\right)^2 = f(y) \, \left( f(x)\right)^2$. Thus if neither $f(x)$ not $f(y)$ equal $0$ you see that $f(y)=f(x)$. Since $f$ is continuous this implies it is a constant. – lulu Aug 30 '17 at 17:56
• @Gribouillis A priori, all I have shown is that $f$ takes at most two values. It can take $0$ but all the non-zero values must coincide. If we add continuity this means that $f$ is constant. Absent continuity, it could take two values. But of course the functional equation is much more powerful than what I deduce from it, so it is likely that there isn't a discontinuous example. – lulu Aug 30 '17 at 18:02
• @Gribouillis I believe I have shown that $f$ is a constant (without using continuity). See my posted solution below. – lulu Aug 30 '17 at 18:09

in fact, the assumptions imply that $f$ is a constant.

proof: Suppose there is some $x_0$ with $f(x_0)=0$. Then for any $x$ we have $$f(x)=f(x-x_0+x_0)=f(x-x_0)\,\left(f(x_0)\right)^2=0$$ so in this case the function is identically $0$.

Alternatively suppose $f(x)\neq 0\,\, \forall x$. Then take two values $x,y$. We have $$f(x+y)=f(y+x)\implies f(x)\,\left( f(y) \right)^2 = f(y)\,\left( f(x) \right)^2\implies f(x)=f(y)$$

and we are done.

Note: Suppose that $f(x)=c$ for some constant $c$ Then we have $$c=f(x)=f(x+0)=f(x)\, \left( f(0) \right)^2=c^3$$ Thus $c=c^3$ so $c=0,\pm 1$. Any of these are possible.

• $c$ can also be $-1$ – Surb Aug 30 '17 at 18:12
• @Surb Thanks, absolutely right. I have edited my post to reflect that. – lulu Aug 30 '17 at 18:14
• So you proved that the statement is true without assumption that $f$ is differentable at $0$. – Aqua Aug 30 '17 at 19:01
• @johnnobody Yes. It's a very strong functional equation. – lulu Aug 30 '17 at 19:08
• Very simple and beautiful answer. +1 Without using any analytic properties you show that the functional equation itself makes the function a constant. – Paramanand Singh Aug 31 '17 at 2:19

If we put $x=y=0$ we get $f(0)=f(0)^3$ so $f(0)\in \{0,1,-1\}$.

a) So if $f(0)=1$ we have

\begin{eqnarray*} \lim_{h\rightarrow0}\frac{f(x)f(h)^2-f(x)}{h} &=& \lim_{h\rightarrow0}\frac{f(x)(f(h)^2-1)}{h}\\ &=& f(x)\cdot\lim_{h\rightarrow0}\frac{(f(h)-1)(f(h)+1)}{h}\\ &=& f(x)\cdot \lim_{h\rightarrow0}\frac{f(h)-f(0)}{h}\cdot \lim_{h\rightarrow0}(f(h)+1)\\ &=& f(x)f'(0)(f(0)+1)\\ &=& 2f(x)f'(0) \end{eqnarray*}

b) If $f(0)=-1$ we have

\begin{eqnarray*} \lim_{h\rightarrow0}\frac{f(x)f(h)^2-f(x)}{h} &=& \lim_{h\rightarrow0}\frac{f(x)(f(h)^2-1)}{h}\\ &=& f(x)\cdot\lim_{h\rightarrow0}\frac{(f(h)-1)(f(h)+1)}{h}\\ &=& f(x)\cdot \lim_{h\rightarrow0}(f(h)-1)\cdot\lim_{h\rightarrow0}\frac{f(h)-f(0)}{h}\\ &=& f(x)(f(0)-1)f'(0)\\ &=& -2f(x)f'(0) \end{eqnarray*}

c) At last, if $f(0)=0$ then for $y=0$ we get $f(x)= f(x+0)= f(x)f(0)^2=0$ so $f$ is constant and thus differentiable.

Hint $f$ differentiable at $a = 0$ means that for some $\ell$, $$\frac{f(h) - f(0)}{h} \longrightarrow \ell$$

You were proceeding in the right direction. Just note that $$f(x+h)-f(x) =f(x) f(h)^{2}-f(x)f(0)^{2}=f(x)(f(h)+f(0))(f(h)-f(0))$$ and thus the derivative is equal to $f(x) \cdot 2f(0)\cdot f'(0)$ so that the function is differentiable for all $x$.