# Find all differentiable functions $f : [0, +\infty) \rightarrow \mathbb{R}$ such that $f(1) = 1$ and $f(x)f(y) \leq f(xy)$

The exercise

Find all differentiable functions $$f : [0, +\infty) \rightarrow \mathbb{R}$$ such that:

• $$f(1) = 1$$
• $$f(x)f(y) \leq f(xy)$$ for all $$x, y \geq 0$$

My try

I have found that the functions $$x \mapsto x^\alpha$$ for $$\alpha \in [1, +\infty)$$ are solutions. The constant function $$x \mapsto 1$$ is also solution.

As $$f(0)^2 \leq f(0)$$ one also has that $$f(0) \in [0, 1]$$.

If I take a solution $$f$$, my idea would be to try proving it is of the form $$f(x) = x^\alpha$$ for some $$\alpha$$. Necessarily this $$\alpha$$ would be equal to $$f'(1)$$. So I set $$\alpha := f'(1)$$ and considered $$g(x) = \frac{f(x)}{x^\alpha}$$ for $$x > 0$$. One can show $$g$$ is derivable and $$g'(1) = 0$$ and $$g$$ verifies the functional equation (for $$x, y > 0$$). So the goal would be to show that $$g = 1$$, but I've not managed to prove it yet. Any help is welcome folks!

• I gave you the problem as it was given to me, and $\tfrac{1}{x}$ is not differentiable in $0$ btw :) Commented Nov 18, 2020 at 8:55
• An idea ( that perhaps does not works). You have $f(xy)-f(x)\geq f(x)(f(y)-1)=f(x)(f(y)-f(1))$. So a) and b): a :Divide by $y-1$ for $y>1$, You get $x\frac{f(x(y-1)+x)-f(x)} {x(y-1)} \geq f(x)\frac{f(y)-f(1)}{y-1}$. Let now $y\to 1$ b: Divide by $y-1$ for $y<1$, and let $y\to 1$. Commented Nov 18, 2020 at 8:59

First note that $$f(1)=1$$, and $$f$$ differentiable, implies that continuous, and hence $$f(x)>0$$, in $$(1-\delta,1+\delta)$$, for some $$\delta>0$$.
Next, if $$x\in (0,\infty)$$, then $$x_n=x^{1/n}\in(1-\delta,1+\delta)$$, for $$n$$ sufficiently large, and hence $$f(x_n\cdot x_n)\ge \big(f(x_n)\big)^2\quad\Longrightarrow\quad f(x_n\cdot x_n\cdot x_n)\ge f(x_n\cdot x_n)f(x_n)\ge \big(f(x_n)\big)^3 \\ \quad\Longrightarrow\quad f(x)\ge f(x_n^n)\ge\big(f(x_n)\big)^n>0.$$ Thus $$f(0,\infty)\subset (0,\infty)$$.
Set next $$g(x)=\log f(\mathrm{e}^x)$$. Then $$g:\mathbb R\to\mathbb R$$, $$g(0)=0$$ and $$g(x+y)\ge g(x)+g(y) \quad \text{for all x,y\in\mathbb R.}$$ The $$g$$ is also differentiable as a composition of such and let $$c=\lim_{h\to 0} \frac{g(h)-g(0)}{h}=\lim_{h\to 0} \frac{g(h)}{h}.$$ Then $$g(x+h)\ge g(x)+g(h) \quad\Longrightarrow\quad \frac{g(x+h)-g(x)}{h}-\frac{g(h)}{h}= \left\{\begin{array}{ccc}\text{non-negative if h>0}, \\ \text{non-positive if h<0.} \end{array}\right.$$ and since $$\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}$$ exists, then $$\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}=\lim_{h\to 0}\frac{g(h)}{h}=c$$ which implies that $$g(x)=cx$$ and $$f(x)=\mathrm{e}^{g(\log x)}=\mathrm{e}^{c\log x}=x^c.$$
Finally, as $$f$$ is defined and it is also differentiable at $$x=0$$, then $$c\ge 1$$.