Finite dimensional dual space

Question

Let $X$ be a normed linear space with a finite dimensional dual $X^*$. How do I prove X is also finite dimensional?

score 1 · Accepted Answer · answered Feb 01 '20 at 21:14

1

If $X^*$ is finite dimensional then so is $X^{**}$. Also, there is an embedding $X\subseteq X^{**}$ by $x\to \phi_x$ where $\phi_x(f)=f(x)$. A subspace of a finite dimensional space is finite dimensional.

answered Feb 01 '20 at 21:14

Mark

39,605

Why is there an embedding $X \subseteq X^{**}$? – nonpara Feb 02 '20 at 00:36
This is the embedding I described. Define a function $i:X\to X^{**}$ by $i(x)=\phi_x$ where $\phi_x(f)=f(x)$. It can be checked that this is an injective linear transformation. – Mark Feb 02 '20 at 00:43

score 0 · Answer 2 · answered Feb 01 '20 at 21:30

0

If $\{f_i\}\quad i=1,2...n$

is a basis of $V^*$, than the set

$\{v_j\}\quad j=1,2...n$ such that $f_i(v_j)=\delta_{ij}$

is a basis for $V$

for $x\in V$ we have:

$x=\sum_{k=1}^nf_k(x)v_k$

answered Feb 01 '20 at 21:30

Emilio Novati

62,675

Why is ${v_i}$ a basis for $V$? – nonpara Feb 02 '20 at 00:37
Because, as you se in the last equation, any Victor can be e expressed as a linear combination of vectors in $V$ – Emilio Novati Feb 02 '20 at 19:34

Finite dimensional dual space

2 Answers2