If $f:\mathbb R\to\mathbb R$ is "almost-linear" so $\{f(x+y)-f(x)-f(y)\}$ is bounded, does $f$ map bounded sets to bounded sets?

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If a function $f:\mathbb R\to\mathbb R$ is "almost-linear", in the sense that

$$\exists M>0:\forall x\in\mathbb R:\forall y\in\mathbb R:|f(x+y)-f(x)-f(y)|\leq M,$$

does it follow that

$$\forall L>0:\exists N>0:\forall x\in\mathbb R:|x|\leq L\implies|f(x)|\leq N\;?$$

Two functions being boundedly close to each other is an equivalence relation. Almost-linearity says that $f(x+y)$ is boundedly close to $f(x)+f(y)$. Taking $y=nx$ and using induction, we can see that, for any fixed $n\in\mathbb Z$, $f(nx)$ is boundedly close to $nf(x)$. Rationally rescaling $x$ to $\tfrac mnx$, we can further see that $f(\tfrac mnx)$ is boundedly close to $\tfrac mnf(x)$; but the "distance" between these functions may depend on $\tfrac mn$....

Related, but I'm not assuming that $f$ is continuous.

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No. There are discontinuous functions $f$ such that $f(x+y)=f(x)+f(y)$ for all $x,y$ and these are not bounded in any open interval around $0$.