Let $g:([0,1])\to([0,1])$ be uniformly continuous. Consider the metric space $(C[0,1];[0,1]), ||. ||_{\infty})$

Show that $G: C([0,1];[0,1])\to C([0,1];[0,1]), f \mapsto g \circ f$ is continuous.

First question, under which metric is $g$ uniformly continuous? Surely the fact that $(C[0,1];[0,1]), ||. ||_{\infty})$ is given does not mean that we are looking at $(g, ||. ||_{\infty})$, or am I wrong?

I was able to solve for $f \mapsto f \circ g$ but I am stuck on the above mentioned variation.

Any help is greatly appreciated.


1 Answer 1


I am not sure I understand your notation correctly. But it seems to me, $g$ is a real-valued function from $[0,1]$ to $[0,1]$, and the metric space $(C([0,1],[0,1]), \|\cdot\|_\infty)$ is the space of real-valued continuous functions from $[0,1]$ to $[0,1]$ equipped with the uniform metric (or supremum metric).

For your first question, you are correct, $g$ should be measured by the 'usual' metric, I mean the metric used by real numbers. Applying the supremum metric on real numbers makes no sense.

If my above interpretation is correct, the proof is quite straightforward. To show $f \mapsto g\circ f$ is continuous under the uniform metric, we need to show $\forall\epsilon>0$, $\exists \delta>0$, such that $$\sup_{x\in[0,1]} |f_1(x) - f_2(x)| < \delta \Rightarrow \sup_{x\in[0,1]} |g\circ f_1(x) - g\circ f_2(x)| < \epsilon \tag{*} $$

Since $g$ is uniformly continous, there exists $\delta>0$ such that for all $x,x'$ with $|x-x'|<\delta$, we have $|g(x) - g(x')| < \epsilon$. We could just take this $\delta$, then the condtion of $(*)$ is satisfied.


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