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I am looking for an example (with proof) of two norms defined on the same vector space, such that the norms on the two spaces are NOT equivalent, but such that one norm dominates the other...

Davide Giraudo
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Frank
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3 Answers3

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Take for instance the space $\mathcal C([0,1])$ of continuous functions on $[0,1]$. The $\mathbb L^2$ norm is dominated by the uniform norm but they are not equivalent, since convergence in $\mathbb L^2$ does not imply uniform convergence.

roger
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Let $(X,\lVert \cdot\rVert)$ be an infinite dimensional normed space and $f\colon X\to \Bbb R$ be a linear non continuous form (for the mentioned norm). Define $N(x):=\lVert x\rVert+|f(x)|$; then $N$ is a norm which is not equivalent to $\lVert\cdot\rVert$.

Davide Giraudo
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For $f\in L^2((0,1))$, we have, by Cauchy-Schwarz: $$ \int_0^1|f(x)|dx\leq \sqrt{\int_0^1|f(x)|^2dx}. $$ In particular, $L^2((0,1))$ is contained in $L^1((0,1))$.

Now if the $L^2$ norm was dominated by the $L^1$ norm, on $L^2(0,1)$, we would have in particular, for $f_n(x)=\frac{1}{\sqrt{x}}1_{(1/n,1)}$, $$ \sqrt{\log n}=\sqrt{\int_\frac{1}{n}^1\frac{1}{x}dx}\leq C\int_\frac{1}{n}^1\frac{1}{\sqrt{x}}dx\leq 2C\qquad \forall n\geq 1. $$

So $L^2((0,1))$ is a counterexample, with the $L^1$ and the $L^2$ norms.

Note: in short, $L^2((0,1))$ is not complete for the $L^1$ norm.

Julien
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