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I know that all norms on a finite dimensional vector space induce the same topology. Moreover, I know that there are infinite dimensional vector spaces with norms that don't induce the same topology. But I am curious about the existence of two infinite dimensional vector spaces:

(1) Is there any infinite dimensional vector space such that all norms induce the same topology? (2) Is there an infinite dimensional vector space such that no two norms induce the same topology?

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  • $\begingroup$ (2) is easy: No. If $\| \cdot \|$ is a norm, then $2\| \cdot \|$ is an equivalent norm, producing the same topology. $\endgroup$ – Theo Bendit Apr 6 '19 at 21:23
  • $\begingroup$ @TheoBendit Aw, shoot. Of course! Thanks for the input! $\endgroup$ – user193319 Apr 6 '19 at 21:24
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Question (1): No.

You didn't mention which field you consider, so let us assume that all vector spaces are vector spaces over $\mathbb R$.

Let $V$ be a vector space with an infinite basis $\mathcal B$. Let $\mathcal B _0 \subset \mathcal B $ be countably infinite subset and $V_0 \subset V$ the subspace spanned by $\mathcal B _0$, $V'$ be the subspace spanned by $\mathcal B \setminus \mathcal B _0$. It is possible that $V' = \{ 0 \}$. Then $V = V_0 \oplus V'$. Take any norm on $V'$. Then any two norms $\lVert - \rVert_i$ on $V_0$ induce two norms on $V$, and these are equivalent (i.e. induce the same topology on $V$) iff the norms $\lVert - \rVert_i$ are equivalent. This reduces your question to showing that $V_0$ has non-equivalent norms. It suffices to consider $\ell_0 =$ set of real sequences $x = (x_n)$ such that $x_n = 0$ for almost all $n$. Let $\lVert x \rVert_1 = \sum \lvert x_i \rvert$ and $\lVert x \rVert_2 = \max \lvert x_i \rvert$. Let $x^n = (1,2,\dots,n-1,n,0,\dots)$. Then $\lVert x^n \rVert_1 = (n+1)n/2$ and $\lVert x^n \rVert_2 = n$, i.e. $\lVert x^n \rVert_1 = (n+1)/2 \cdot \lVert x^n \rVert_2$. This shows that $\lVert - \rVert_1, \lVert - \rVert_2$ are not equivalent.

Question (2): No.

See the comments.

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On every infinite dimensional normed linear space there is a discontinuous linear functional. If $f$ is one such functional then $\|x\|+|f(x)|$ is another norm and this norm is not equivalent to $\|x\|$. Part 2) is already answered above.

Construction of $f$: let $H$ be a Hamel basis for the space and pick a sequence of distinct points $h_1,h_2,...$ in $H$. Let $f(h_n)=n\|h_n\|,n=1,2...$ and $f(x)=0$ for any $x \in H$ which is not one of the vectors $h_n$. Extend $f$ to the whole space by linearity. Then $f$ is a linear functional which is not continuous.

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  • $\begingroup$ First sentence needs reference. $\endgroup$ – GEdgar Apr 6 '19 at 23:42
  • $\begingroup$ @GEdgar Since the argument is quite simple I have included a proof . $\endgroup$ – Kavi Rama Murthy Apr 7 '19 at 12:28

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