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Two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are said to be equivalent if there exists constant $K>0$ and $M>0$ such that $$K\|x\|_1\le\|x\|_2\le M\|x\|_1,\quad x\in\mathbb{R}^n$$.

Let denote the norm space $\mathbb{R}^n$ by $(\mathbb{R}^n,\|\cdot\|)$ equipped with the norm $\|\cdot\|$. A map $f:(\mathbb{R}^n,\|\cdot\|)\to\mathbb{R}$ is uniform continuous if $$\forall\,\epsilon>0\,\,\exists\,\delta>0\,\ni y,x\in \mathbb{R}^n\,\,\|x-y\|<\delta\,\,\Longrightarrow\,\,|f(x)-f(y)|<\epsilon$$

I wish to show $f$ is uniformly continuous w.r.t. the norm $\|\cdot\|_1$ if and only if it is uniformly continuous w.r.t. the norm $\|\cdot\|_2$.

We know that all norms on $\mathbb{R}^n$ are equivalent. So we have $K\|x\|_2\le \|x\|_1\le M\|x\|_2$, so how can we utilise this result to prove the above claim?

Any help is appreciated.

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  • $\begingroup$ You get $\delta_1$ for norm 1. How would you select $\delta_2$ for norm 2? $\endgroup$ – user251257 Oct 7 '15 at 14:01
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Suppose $f$ is uniformly continuous w.r.t $||\cdot||_1$.

Using $K||x||_1 \le ||x||_2 \le M||x||_1$.

Let $\epsilon >0$.

For this you can find $\delta_1 >0$ such that, $ \forall \ x,y \in \mathbb{R}^n, ||x-y||_1< \delta_1 \implies |f(x)-f(y)|< \epsilon$

Now use the norm equivalence relation, $||x-y||_2 \le M \ ||x-y||_1 < M \delta_1 $.

So, for the given $\epsilon$ take $\delta_2= M \delta_1$. This $\delta_2$ meets the requirement for uniform continuity of $f$ w.r.t. $|| \cdot||_2$.

$\forall\ \epsilon >0,\ \exists\ \delta_2 >0 $ such that $\forall x,y \in \mathbb{R}^n, ||x-y||_2< \delta_2 \implies |f(x)-f(y)|< \epsilon$.

Hence f is uniformly continuous w.r.t. $|| \cdot||_2$.

Similarly you can prove the converse by taking $\delta_1= \frac{\delta_2}{K}$.

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  • $\begingroup$ Let me rephrase what you had, $\forall\epsilon>0\,\exists\delta_1>0\,\ni\,\forall\,x,y\in\mathbb{R}^n,\, \|x-y \|_2<\delta_1\,\Longrightarrow\,|f(x)-f(y)|<\epsilon$. Then via the norm equivalence relation (the version I wrote), we have $\|x-y\|_1\le M\|x-y\|_2<M\delta_1$, for the rest, I'm sorry but I do not understand... $\endgroup$ – math101 Oct 8 '15 at 7:36
  • $\begingroup$ I also noted that the standard version of norm equivalence is $K\|x\|_1\le \|x\|_2\le M\|x\|_1$. But in my own proof, I get what I wrote above about the norm equivalence result. $\endgroup$ – math101 Oct 8 '15 at 7:41
  • $\begingroup$ Both the versions are the same. Its just a different way of representing it. Which one do you want me to use in the proof? It can be proved using either one of them. $\endgroup$ – Epsilon Oct 8 '15 at 7:46
  • $\begingroup$ i think i got your point, let me rewrite so that you can correct me. $\endgroup$ – math101 Oct 8 '15 at 8:03
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You note that $||f (x)-f (y)||_1 <M\epsilon \Longrightarrow ||f (x)-f (y)||_2 <\epsilon $ and $ ||f (x)-f (y)||_2 <K\epsilon \Longrightarrow ||f (x)-f (y)||_2 <\epsilon $.

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  • $\begingroup$ No, I am sorry but I do not understand your answer. Lost from what you wrote at the very beginning. Why $\|f(x)-f(y)\|_1<M\epsilon$? $\endgroup$ – math101 Oct 8 '15 at 6:58
  • $\begingroup$ $f(x),f(y)$ are elements in $\mathbb{R}$. There won't be any $||.||_1$ or $||.||_2$ around them. It should be $|f(x)-f(y)|$. $\endgroup$ – Epsilon Oct 8 '15 at 7:07
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The idea is to work from one side, the uniform continuity of norm 1 yields an arbitrary $\delta_1,$ then invoke the norm equivalence result, we found the choice of $\delta$ works for norm 2.

Let $f$ be uniformly continuous w.r.t. $\|\cdot\|_1$. Then, $$\forall\,\epsilon>0\,\exists\delta_1>0\,\ni\,\forall \ x,y\in\mathbb{R}^n,\,\|x-y\|_1<\delta_1\,\,\Longrightarrow\,\,|f(x)-f(y)|<\epsilon.$$

Now, via the norm equivalence relation in $\mathbb{R}^n$, we have $$K \|x-y\|_2\le\|x-y\|_1$$ $$\Longrightarrow\,\|x-y\|_2\le\frac{1}{K}\|x-y\|_1<\frac{1}{K}\delta_1.$$

Let $\delta_2=\frac{1}{K}\delta_1$. So we have found such a $\delta_2$, which meets the requirement for uniform continuity of $f$ w.r.t. $\|\cdot\|_2,$ i.e.

$$ \forall\epsilon>0\ \exists\ \delta_2\ni\forall x,y\in \mathbb{R}^n \|x-y\|_2<\delta_2 \Longrightarrow |f(x)-f(y)|<\epsilon.$$ .

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  • $\begingroup$ It is correct.. $\endgroup$ – Epsilon Oct 8 '15 at 8:43

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