How to prove for if every normed vector space $\mathrm E$ is infinite dimensional then its dual $\mathrm E^*$ is infinite dimensional as well.

I know that If $\mathrm E$ is a finite dimensional vector space, we can identity $\mathrm E$ with $\Bbb K^n$ as vector space and hence the norm $\vert\vert.\vert\vert_2$ on $\mathrm E$. If $\vert\vert.\vert\vert$ is an norm on $\mathrm E$, then $\vert\vert.\vert\vert$ and $\vert\vert.\vert\vert_2$ are equivalent.

How about infinite dimensional normed vector space?

  • $\begingroup$ Can you use a basis for $E$ to define a basis for $E^*$? $\endgroup$ – Theoretical Economist Dec 9 '16 at 1:17
  • 2
    $\begingroup$ If $E^*$ were finite dimensional so would be $E^{**}$ which contains a copy of $E$. $\endgroup$ – Jochen Dec 9 '16 at 8:27

Prove it by contraposition. Suppose that $F$ is finite-dimensional. Then so is $F^*$ as the coordinate functionals associated to any (finite) basis of $F$ span $F^*$. On the other hand, if $E$ is infinite-dimensional, then it is isomorphic to a subspace of $E^{**}$ via $\kappa_E\colon E\to \kappa_E(E)\subseteq E^{**}$ given by $$\langle \kappa_E x, f\rangle = \langle f,x\rangle\quad (x\in E, f\in E^*). $$

Subspaces of finite-dimensional spaces are finite-dimensional.

Note that the result is no longer true for more general topological vector spaces. For example $L_p[0,1]^*=\{0\}$ for $p\in (0,1)$.


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