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I'm doing Exercise 1.8.1 in Brezis's book of Functional Analysis. Could you verify if my attempt is fine?

Let $E$ be an n.v.s. with norm $|\cdot|$. Let $C \subset E$ be open convex such that $0 \in C$. Let $p$ denote the gauge of $C$. Assuming $C$ is symmetric (i.e., $-C=C$ ) and $C$ is bounded, prove that $p$ is a norm which is equivalent to $|\cdot|$.

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Analyst
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  • see also https://math.stackexchange.com/questions/1495726/norm-induced-by-convex-open-symmetric-bounded-set-in-bbb-rn – daw Apr 21 '22 at 11:36

1 Answers1

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We have $$p(x) := \inf \{ r > 0 \mid x/r \in C\} \quad \forall x\in E.$$

It has been proved in the textbook that

  • $p$ is sublinear.
  • there is a constant $M$ such that $0 \leq p(x) \leq M |x|$ for all $x \in E$.
  • $C=\{x \in E \mid p(x)<1\}$.

Because $C$ is symmetric, $x\in C \iff -x \in C$. This implies $$ p(x) = \inf \{ r > 0 \mid x/r \in C\}=\inf \{ r > 0 \mid -x/r \in C\} = p(-x). $$

Hence $p(\lambda x) = p(-\lambda x) = \lambda p(x)$ for all $\lambda \ge 0$. This implies $p(\lambda x) = |\lambda| p(x)$ for all $\lambda$. Assume $C$ is bounded by $R>0$. We have \begin{align} p(x) = 0 & \iff \forall r>0, \exists 0<r'<r: x/r'\in C \\ & \iff \forall r>0, \exists 0<r'<r: x \in r'C \\ & \implies \forall r>0, \exists 0<r'<r: |x| \le r' R \\ & \implies |x|=0 \iff x=0. \end{align}

Hence $p(x)=0 \iff x=0$. Thus $p$ is indeed a norm on $E$. For $r >0$ such that $x/r \in C$, we have $x \in r C$ and thus $|x| \le rR$. It follows that $p(x) \ge |x|/R$. Then we have $|x|/R \le p(x) \le M|x|$ for all $x\in E$. This implies $p$ is equivalent to $|\cdot|$.

Analyst
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