Uniform continuity of composition?

Question

If we consider $f\circ u$ and know that

f is bounded and uniformly continuous
u is bounded and continuous

does this imply that $f\circ u$ is bounded and uniformly continuous?

It is clear that it is bounded.

But it is not clear to me whether it is uniformly continuous. Surely it is continuous.

score 0 · Accepted Answer · answered Jun 25 '20 at 14:17

0

The answer is no. You can give the following counter-example. Let your domain be $(0,1)$, $f(x)= x$ and $g(x)= x^2$. Then $f(g(x))= f(x^2)= x^2$ which is bounded, continuous, but not uniformly continuous on the given domain.

answered Jun 25 '20 at 14:17

Davi Barreira

3,248

What if $u\in C([0,T]\times\mathbb{R})$ where $T>0$? Does this make $f(u)$ uniformly continuous? – Salamo Jun 25 '20 at 17:12
Yes, because the domain would be compact, hence $u$ is uniformly continuous and one can prove that composition of uniformly continuous functions is also uniformly continuous – Davi Barreira Jun 25 '20 at 17:17
I hope this is not too stupid: But why is the domain compact in this case? – Salamo Jun 25 '20 at 17:18
Perhaps I misunderstood the notation, $u \in C([0,T], \mathbb R)$ doesn’t imply that $u:[0,1]\rightarrow\mathbb R$? – Davi Barreira Jun 25 '20 at 17:20
No, the notation means that $u\colon [0,T]\times\mathbb{R}\to\mathbb{R}$ and $u$ is continuous. – Salamo Jun 25 '20 at 17:21
I can’t answer right away, would have to give some thought – Davi Barreira Jun 25 '20 at 17:26
I'll ask this as a separate question. – Salamo Jun 25 '20 at 17:39

Uniform continuity of composition?

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