# If any two norms on a vector space are equivalent then the space is finite-dimensional [duplicate]

I need to prove: If any two norms on a vector space are equivalent then the space is finite-dimensional.

I am aware of the converse of this result that on a finite dimensional vector space any two norms are equivalent. Any kind of hint is appreciated.

• Do you know that any infinite-dimensional normed vector space admits an unbounded linear functional? – PhoemueX Sep 18 '14 at 6:49
• No. How to construct that? – QED Sep 18 '14 at 7:03
• – PhoemueX Sep 18 '14 at 7:06

HINT:

Let $X$ an infinite dimensional real vector space and let $(e_i)_{i\in I}$ be a basis of $X$ as a vector space.

Let $w\colon I \to (0, \infty)$, $i \mapsto w_i$ a function (the weights) such that both $(w_i)_{i\in I}$ and $(1/w_i)_{i \in I}$ are not bounded from above. This is possible since $I$ is infinite.

Consider the norms $||\cdot ||_1$ , $||\cdot ||_2$ on $X$ defined by:

\begin{eqnarray} ||\sum a_i e_i ||_1 \colon &= &\sum_i |a_i| \\ ||\sum a_i e_i||_2 \colon& = &\sum_i w_i |a_i| \end{eqnarray}

The norms $||\cdot ||_1$ , $||\cdot ||_2$ are not comparable, and therefore are not equivalent.

• Instead of $\| {}\cdot {}\|_2$ which involves $w$, you can just consider $\| \sum a_i e_i \|_{\infty} = \max_i |a_i|$ (note that all but finitely many $a_i$ must be zero). Then, for any positive integer $N$, if $x_N=\sum_{i=1}^{N} e_i$, we have $\|x_N\|_1 = N = N\| x_N \|_{\infty}$.This shows we cannot have $\| {}\cdot {}\|_{1} \leq C\| {}\cdot {}\|_{\infty}$, hence the norms are inequivalent. – MichaelGaudreau Sep 25 '18 at 23:19