# Is the family of equivalent norms a locally compact space?

Consider an infinite dimensional Banach space $$(X,||\cdot||)$$. Let $$\mathcal{P}$$ be the family of all equivalent norms in $$X$$. That is $$p\in \mathcal{P}$$ iff $$p$$ is equivalent to $$||\cdot||$$, i.e. there exists two constants $$c_1,c_2>0$$ such that $$c_1||x||\leq p(x)\leq c_2||x||$$ for all $$x\in X$$.

$$\mathcal{P}$$ is a metric space endowed with the following metric $$\rho$$:

$$\rho(p,q)=\sup_{||x||\leq 1} \lbrace |p(x)-q(x)|\rbrace$$

$$\mathcal{P}$$ is a Baire space, seen as an open subset of the space of all continuous semi-norms in $$(X,||\cdot||)$$ with the same metric $$\rho$$. This last space is a complete metric space, so by the Baire category theorem it is a Baire space.

Is $$\mathcal{P}$$ a locally compact space? - We can't use Riesz Lemma, as this space is not a linear space, to argue against the locally compactness.

• what does "the family of all equivalent norms" mean?? you're looking at all norms but consider two to be equal in $\mathcal{P}$ if they are equivalent? – mathworker21 Sep 30 '18 at 10:23
• @mathworker21 An element $p\in \mathcal{P}$ means that it is equivalent to $||\cdot||$, i.e. there exists two constants $c_1,c_2>0$ such that $c_1||x||\leq p(x)\leq c_2||x||$ for all $x\in X$ – Chazz Oct 1 '18 at 4:06

No, here is a counterexample. Take $$X=L^1(\mathbb{R})$$, $$\|.\|_1$$ the usual $$L^1$$ norm.

By contradiction, assume $$\mathcal{P}$$ to be locally compact. Then there exists a compact neighbourhood $$K$$ of $$\|.\|$$ which is compact. Let $$\epsilon>0$$ be such that $$B(\|.\|_1,2\epsilon)$$ is a subset of $$K$$.

Define for $$f \in X$$, $$p_n(f)=\int_{\mathbb{R}} |f(t)|\left(1+\epsilon e^{-(t-n)^2}\right)dt.$$

These norms are equivalent to $$\|.\|_1$$ (with $$c_1=1$$ and $$c_2=1+\epsilon$$).

Clearly, $$\rho(p_n,\|.\|_1)\leq \epsilon$$ and in fact there is equality (by considering a sequence $$(f_k)_k$$ of approximation of unity around $$t=n$$).

Since $$p_n \in K$$ for every $$n$$, one can extract a subsequence converging. However, no extracted sequence of $$(p_n)_n$$ is a Cauchy sequence: for every $$n$$, $$\lim_{m\to +\infty} \rho(p_n,p_m)\geq \epsilon,$$ again by considering a sequence $$(f_k)_k$$ of approximation of unity around $$t=n$$. This yields a contradiction.

• Great! Do you know if there are examples where $\mathcal{P}$ is indeed a locally compact space? – Chazz Oct 1 '18 at 19:19
• I would be surprised if there is, but truth is I don't know. – user120527 Oct 2 '18 at 7:59
• I was wondering if this argument can be modified to actually prove that for all Banach spaces $(X, ||\cdot||)$, the family of all equivalent norms is NOT locally compact? – Chazz Oct 5 '18 at 0:04