Derivatives of functions satisfying Euler-ish inequality $f(x+y)\le f(x)f(y)$.

Question

Conjecture. Let $f:\mathbb R\to\mathbb R_{>0}$ be a continuously differentiable function such that $$f(x+y)\le f(x)\cdot f(y)$$ for all reals $x,y$. Then $$f'(x)\le c_1\cdot\exp(c_2\cdot x)$$ for some constants $c_1,c_2$; and all $x\in\mathbb R$.

Is this conjecture true? How can we (dis-)prove it?

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Remarks

• My conjecture originates from the fact that all continuous functions satisfying $g(x+y)=g(x)\cdot g(y)$ are of the type $g(x)=\exp(c\cdot x)$. In particular, $g'(x)=c\exp(cx).$

• My first instinct was to use $$f(x+y)-f(x)\le f(x)\cdot(f(y)-1)$$ so if we could conclude that $f(x)\le \exp(cx)$ for all $x$, then $$f'(x)=\lim_{y\to0}\frac{f(x+y)-f(x)}{y}\le \exp(cx)\lim_{y\to0}\frac{\exp(cy)-1}y=\exp(cx)\cdot c.$$ However, $f(x)\le \exp(cx)$ is not always true (consider $f(x)=2\cdot\exp x$).

Answer 1

This is unfortunately not true. The issue is that your condition essentially only restricts the behaviour of $f$ "in the large", but not locally. The derivative of $f$, however, is determined by the local behaviour. What exactly I mean by this will be made clear by the counterexample.

Define $$ f : R \to (0,\infty), x \mapsto e^x \cdot (4 + \sin(e^{x^{2}})) . $$ Since $-1 \leq \sin(y) \leq 1$, it is then easy to see $3 e^x \leq f(x) \leq 5 \cdot e^x$ for all $x \in \Bbb{R}$, and hence $$ f(x+y) \leq 5 \cdot e^{x+y} \leq 3 e^x \cdot 3 e^y \leq f(x) \cdot f(y). $$ However, we have $$ f'(x) = e^x \cdot (4 + \sin(e^{x^{2}})) + e^x \cdot \cos(e^{x^{2}}) \cdot e^{x^{2}} \cdot 2x. $$ Now, for $n \geq 5$, let $x_n := \sqrt{\ln (2 \pi n)}$. Observe $\ln(2 \pi n) \geq \ln(e) = 1$, and thus $e^{x_n} \geq e \geq 1$. Furthermore, $e^{x_n^2} = e^{\ln(2 \pi n)} = 2\pi n$, and thus $\cos(e^{x_n^2}) = \cos(2\pi n) = 1$. Overall, this implies $$ f'(x_n) \geq 3 \cdot e^{x_n} + e^{x_n} \cdot \cos(e^{x_n^2}) \cdot e^{x_n^2} \cdot 2 x_n \geq 2 \pi n \cdot 2 x_n \geq 4 \pi n. $$ However, if your desired inequality was true, we would have $$ 2 e^{x_n^2} = 4 \pi n \leq f'(x_n) \leq C \cdot e^{c x_n}, $$ and thus $$ e^{x_n^2 - c x_n} \leq C / 2, $$ meaning that $x_n^2 - c x_n = x_n \cdot (x_n - c)$ is bounded, which is clearly not the case.