# The Functional Inequality $f(x) \ge x+1$, $f(x)f(y)\le f(x+y)$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function that satisfies the following conditons.

$$f(x)f(y)\le f(x+y)$$ $$f(x)\ge x+1$$

What is $f(x)$?

It is not to difficult to find that $f(0)=1$.

If $f(x)$ is differentiable, we can further these results so that $f'(0)=1$, and $f(x)=f'(x)$.

However, I was not able to go any further than this. I believe that $f(x)=e^x$, but cannot prove it.

Any help would be appreciated.

• Maybe you could use Taylor's expansion? – GoodDeeds Jan 26 '16 at 9:31
• Could you show us the proof of $f'=f$ in the case of differentiability? Maybe if you do it carefully, you can actually prove the differentiability instead of assuming it? – John Steinbeck Jan 26 '16 at 9:41

Let's try to get bounds on $f(1)$, for the sake of generalization to $x\in\mathbb{R}$. We can get a nice lower bound by doing the following algorithm: \begin{align} f(\tfrac12+\tfrac12)&\geq f(\tfrac12)^2\geq (\tfrac12+1)^2=\tfrac94\\ f(\tfrac14+\tfrac14+\tfrac14+\tfrac14)&\geq f(\tfrac14)^4\geq (\tfrac14+1)^4=\tfrac{625}{64} \end{align} Note that I did $f(\frac14+\frac14+\frac14+\frac14)\geq f(\frac14+\frac14)^2\geq f(\frac14)^4$. We can continue this process to get for any $k\geq1$ the inequality $$f(1)\geq (\frac{1}{2^k}+1)^{2^k}$$ and it is widely known that this limit goes to, as you conjectured, $e$. We can generalize this by $$f(x)=f(\tfrac{x}{2^k}+\tfrac{x}{2^k}+\cdots+\tfrac x{2^k})\geq f(\tfrac x{2^k})^{2^k}\geq (\tfrac x{2^k}+1)^{2^k}$$ and the last expression will tend to $e^x$ as $k$ approaches infinity. Thus, we proved (notice that we use $\lim f(a_n)=f(\lim a_n)$ because $f$ is continuous) that $f(x)\geq e^x$. Now we're almost done. Since we have $1=f(x)f(-x)$, we know that $f(x)=\frac{1}{f(-x)}\leq \frac{1}{e^{-x}}=e^x$, and thus, $f(x)=e^x$.
• Yes, I think that It is correct. I think that you can also say that for general $n$, not just $2^k$. – Chad Shin Jan 26 '16 at 13:24
• last statement may be in this way: $f(-x)=\dfrac{1}{f(x)} \le e^{-x} \implies f(x) \le e^x \implies f(x)=e^x$ – chenbai Jan 27 '16 at 0:49