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In Analysis on Manifolds, Munkres makes the statement that for most purposes, the sup metric, which is $$max\{|x_{i} - y_{i}| \space i \in \{1 \dots n\}\}$$ and the euclidean metric are "equivalent." In other words, they can be interchanged in theorems like this:

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I cannot find an instance when interchanging the metrics in a theorem is unacceptable -- I suspect that this has to do with the fact that one can place, within any neighborhood of one metric in $\mathbb{R^{n}}$, another neighborhood of the second metric.

Is my hunch correct -- that since neighborhoods formed using either metric can be placed within each other, all theorems involving a metric can be formulated using either metric? At least with respect to analysis on $\mathbb{R^{n}}$?

Muno
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If you are talking about the exponent $\alpha$ of a Hölder condition on a function, then the metric matters: the Hölder exponent of a function can change when you change the metric. Even worse, it might satify a Hölder condition with respect to one metric but no Hölder condition at all with respect to another.

The same issue holds for a Lipschitz condition on a function as well.

Lee Mosher
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  • How can this be true for metrics induced by equivalent norms, as in the question, and as for all norms on $\mathbb R^n$? If $|f(x)-f(y)|\leq C |x-y|_1^\alpha$ for all $x$ and $y$, and $|x-y|_1\leq K|x-y|_2$ for all $x$ and $y$, then $|f(x)-f(y)|\leq CK|x-y|_2^\alpha$ for all $x$ and $y$ (and similarly in the other direction). – Jonas Meyer Feb 14 '17 at 03:24
  • My answer is rather too broad, I was writing about interchange of any two metrics, not just interchange of the sup and euclidean metrics. However, one part of my answer does apply to the sup and Euclidean metrics: the Lipschitz constant of a function can change when you interchange the sup and euclidean metrics. – Lee Mosher Feb 14 '17 at 13:18
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Yes - actually, you have in $\mathbb{R}^n$ that all the norms are equivalent, that means, exists $a, b>0$ such that

$$ a|x|_1\leq |x|_2\leq b|x|_1 $$

This is the same as your neighborhood image - for every sphere (neighborhood using euclidean norm) you can place inside a cube (neighborhood using sup norm). This makes a lot (sadly, not all) of the theorems have very similar proofs when you change the norm.

For example, lets call $|\cdot|_1$ the sup norm, and $|\cdot |_2$ the euclidean norm. If we know that

4.7-Sup: Let $X$ be a compact subspace of $\mathbb{R}^m$; let $f:X\to \mathbb{R^n}$ be continuous. Given $\varepsilon>0$, there is a $\delta>0$ such that whenever $x, y\in X$,

$$|x-y|_1<\delta \text{ implies } |f(x)-f(y)|_1<\varepsilon $$

Using that, we can prove the following theorem

4.7-Euc: Let $X$ be a compact subspace of $\mathbb{R}^m$; let $f:X\to \mathbb{R^n}$ be continuous. Given $\varepsilon>0$, there is a $\delta>0$ such that whenever $x, y\in X$,

$$|x-y|_2<\delta \text{ implies } |f(x)-f(y)|_2<\varepsilon $$

Proof: Recall that exists some (fixed) $a, b>0$ such that, for all $p\in \mathbb{R}^m$

$$a|p|_1\leq |p|_2 \leq b|p|_1$$

Let $\varepsilon>0$. We know that exists a $\eta>0$ such that, for all $x, y \in \mathbb{R^m}$

$$ |x-y|_1<\eta \text{ implies } |f(x)-f(y)|_1<\frac{\varepsilon}{b} $$

Now, if, $\eta> \frac{1}{b}|x-y|_2$, then $\eta>\frac{1}{a}|x-y|_2\geq |x-y|_1$, and then implies $\varepsilon>b|f(x)-f(y)|_1\geq |f(x)-f(y)|_1$. And calling $\delta=\eta b >0$, we have

$$ |x-y|_2<\delta \text{ implies } |f(x)-f(y)|_1<\varepsilon$$

So, 4.7-Sup implies 4.7-Euc.

Nicolás Vilches
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The Euclidean norm is strictly convex, and the sup norm isn't. This is relevant when unique best approximations are relevant. For example, given a closed convex subset $X$ of $\mathbb R^n$, and a point $p\in\mathbb R^n$, there is a unique $x\in X$ with smallest Euclidean distance to $p$. This is not true with the sup norm; e.g., in $\mathbb R^2$, if $X$ is the closed unit "ball" in the sup norm, and $p=(2,0)$, then the smallest distance from $p$ to $X$ is $1$, but this minimum is achieved for all of the infinitely many points $(1,t)\in X$.

Jonas Meyer
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