Franklin's Notes


Dual norm

Given a norm $\lVert\cdot\rVert$ on a vector space $\mathbb F^n$, the corresponding dual norm $\lVert\cdot\rVert '$ is defined as follows:

It can also be interpreted as the matrix norm induced by $\lVert \cdot\rVert$ of the matrix $x^\ast$.

Theorem 1. The Euclidean norm $\lVert\cdot\rVert_2$ is self-dual. That is, $\lVert\cdot \rVert_2'=\lVert\cdot\rVert_2$.

Proof. For any given vector $x\in\mathbb R^n$, $\lVert x\rVert_2'$ is defined as By Cauchy-Schwarz , we know that $|y^\ast x|\leq \lVert x\rVert_2 \lVert y\rVert_2$, meaning that the above supremum is less than or equal to $\lVert x\rVert_2$. However, letting $y=x$ allows it to attain the value $\lVert x\rVert_2$, so the supremum equals this value. Hence, $\lVert x\rVert_2' = \lVert x\rVert_2$. $\blacksquare$

Theorem 2. The dual of the $p$ norm is the $q$ norm, with $p^{-1}+q^{-1}=1$. That is, $\lVert\cdot \rVert_p'=\lVert\cdot\rVert_q$.

Proof. This proof is very similar. For $x\in\mathbb R^n$, $\lVert x\rVert_p'$ is defined as By Hölder's Inequality , we know that $|y^\ast x|\leq \lVert x\rVert_q \lVert y\rVert_p$, meaning that the above supremum is at most $\lVert x\rVert_q$. However, we know that equality is attained in Hölder's Inequality for some $y$, meaning that this supremum actually equals $\lVert x\rVert_q$. $\blacksquare$

We also have the following convenient fact:

Theorem 3. Every norm on $\mathbb F^n$ is the dual of its dual. That is, $\lVert\cdot\rVert '' = \lVert\cdot\rVert$ for all norms $\lVert\cdot\rVert$ on $\mathbb F^n$.

I think the proof of the above theorem requires some topological machinery that I'm not familiar with yet.

linear-algebra

norm

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