# Functional inequality

Let $f, g:\Bbb R \to \Bbb R$ be bounded functions satisfying $$|f(x+y)-f(x)g(y)|\le \frac 1 4$$ for all $x, y\in \Bbb R$.

Prove or disprove $$|f(x)||1-g(y)|\le \frac 1 4$$ for all $x, y\in \Bbb R$.

-

This is not generally true. Counterexample: Take $f(x)=\frac{\sin x}4$ and $g(x)=\cos x$.
Then, $$\left|f(x+y)-f(x)g(y)\right|=\frac14\left|\sin(x+y)-\sin x\cos y\right|= \frac14\left|\sin y\cos x\right|\le \frac14$$ while for $x=\frac{\pi}2$ and $y=\pi$: $$\left|f(\frac{\pi}2)\right|\left|1-g(\pi)\right|=\left|\frac{1}4\right|\left|2\right|=\frac12> \frac14$$