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Fix $p>1$. Let $\mathscr C[0,2]$ be the space of continuous functions on $[0,2]$ with metric given by $$d_{p}(f,g)=||f-g||_p:=\left(\int_0^2 |f(x)-g(x)|^p \ dx\right)^{\frac{1}{p}}.$$ Consider the map $\Psi:\mathscr C[0,2]\rightarrow \mathbb R$ given by $$\Psi(f)=\left(\int_0^2 f(x)\ dx \right)^4.$$

Writing $\Psi = T\circ\Phi$ where $T(x)=x^4$ and $\Phi:\mathscr C[0,2]\rightarrow \mathbb R$ is given by $$\Phi(f)=\int_0^2 f(x)\ dx,$$ one can show that $\Phi$ is uniformly continuous, and so $\Psi$ is continuous. However, one can also show that $\Psi$ is $\textbf{not}$ uniformly continuous: let $f_n(x) = n+\frac{1}{n^3}$ , $g_n(x)= n$. Then $$||f_n-g_n||_p=\frac{2^{1/p}}{n^3}\rightarrow 0 \quad \text{as}\quad n\rightarrow \infty.$$

On the other hand, $$|\Psi(f_n)-\Psi(g_n)|=16\left(\left(n+\frac{1}{n^3}\right)^4-n^4\right)\not\rightarrow 0\quad \text{as}\quad n\rightarrow \infty.$$

This seems to contradict the fact that the composition $f\circ g$ of two uniformly functions $f,g:\mathbb R\rightarrow \mathbb R$ is uniformly continuous (see this for example).

Is this phenomenon due to the different metrics used?

cmk
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Timothy
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    What's contradictory here? $T$ is not uniformly continuous, so we shouldn't expect $\Psi = T \circ \Phi$ to be uniformly continuous, even though $\Phi$ is. – user125932 Oct 14 '19 at 17:50

1 Answers1

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We have to take $\mathbb{R}$ as the input space of $T$, as $\Phi$ can take values in all of $\mathbb{R}$. You were probably thinking about $[0;2]$ as the input space.

As @user125932 already said $T:\mathbb{R}\rightarrow\mathbb{R}$ is not uniformly continuous. As a counterexample consider a fixed $\delta>0$ and $y=x+\delta$: $$\lim_{x\rightarrow \infty}\vert x^4-(x+\delta)^4\vert=\lim_{x\rightarrow \infty}\vert 4x^3\delta+6x^2\delta^2+4x\delta^3+\delta^4\vert=\infty$$ (Another way of showing that $T$ is not uniformly continuous is basicially your way for $\Psi$ as it is the composition of $T$ and a linear function - $\Phi$ restricted to constant functions on $\mathscr{C}[0;2]$ is linear. So thats quite cool aswell! :D)

As for your question: Uniform continuity is compatible with composition and independent of the metrics used.

Consider the diagram of metric spaces and uniformly continuous maps: $$(L,d_L)\overset{f}\longrightarrow (M,d_M)\overset{g}\longrightarrow(N,d_N)$$ Since $g$ is uniformly continuous we can find for any given $\varepsilon>0$ a $\delta>0$ so that for all $x,y\in M$ we have: $$d_M(x,y)<\delta\;\Rightarrow\;d_N\left(g(x),g(y)\right)<\varepsilon$$ As $f$ is also uniformly continuous there always exists a $\delta'$ such that for all $x,y\in L$: $$d_L(x,y)<\delta'\;\Rightarrow\;d_M\left(f(x),f(y)\right)<\delta$$ So in particular for every $\varepsilon>0$ there exists a $\delta'>0$ such that for all $x,y\in L$ we have: $$d_l(x,y)<\delta'\;\Rightarrow\;d_N\left(g(f(x)),g(f(y))\right)<\varepsilon$$ and $g\circ f$ indeed is uniformly continuous. :)

Also I'd like to genuinely say that your question was extremely well written and concise, that doesn't happen that often :P

C. Brendel
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    I first thought we restrict to the interval $[0,2]$ and $T$ is uniformly continuous there. However we apply $T$ to the image of $\Phi$ and that is of course all of $\mathbb{R}_{\ge 0}$ so your argument is correct. Maybe you can point that out in your answer. – quarague Oct 16 '19 at 08:14
  • If fact the image of $\Phi$ is all of $\mathbb{R}$. Still this is a very nice addition, thanks! – C. Brendel Oct 16 '19 at 10:11
  • @C.Brendel Yes, the confusion was the domain of T. Thanks for the clarification! – Timothy Oct 16 '19 at 15:47
  • Ok! Good to know! No problem :) – C. Brendel Oct 16 '19 at 15:53