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How many non-equivalent norms can we define in an infinite dimensional vector space? Is there any explicit expression?

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There are plenty of non-equivalent norms. Let $X$ be an infinite-dimensional normed space with norm $\|\cdot\|_X$.

Let $Y$ be another normed space with norm $\|\cdot\|_Y$. Let $T\in \mathcal L(X,Y)$ be compact and injective. Then $$ \|x\|_T:=\|Tx\|_Y $$ is a norm on $X$. Moreover, $\|x\|_T \le \|T\|_{\mathcal L(X,Y)} \|x\|_X$. However, both norms cannot be equivalent: If there would be a constant $c>0$ such that $$ \|x\|_X \le c\|x\|_T = c\|Tx\|_Y \quad\forall x\in X, $$ this would imply that $T^{-1}$ is a continuous operator from $Im(T)$ to $X$, which is a contradiction (as compact operators cannot have continuous inverses).

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  • $\begingroup$ You say, that there is no constant $c$ such that $c\|x\|_X \ge \|x\|_T$. But you have already proved, that $\| T \|$ is such a constant. It should be $c\|x\|_X \leq \|x\|_T$. Then the inverse of $T$ is bounded, which is impossible. $\endgroup$ – user128245 Jan 21 '17 at 14:26
  • $\begingroup$ Thanks for spotting! I corrected the argument. $\endgroup$ – daw Jan 22 '17 at 13:48
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When you are given just a vector space $X$, then you have as many non-equivalent norms on $X$ as many non-isomorphic normed spaces you can find with the same linear dimension. This is the standard `structure transport' argument. For instance, suppose that you are given a vector space of dimension continuum. Then for each infinite-dimensional Banach space $Y$ of linear dimensional dimension $\mathfrak{c}$ you can find a linear bijection $T_Y\colon X\to Y$ which gives you a norm $\|x\|:=\|T_Yx\|$.

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There are exactly $2^{\dim X}$ inequivalent norms on an infinite dimensional vector space. You can find the proof here.

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