Basic functional analysis question: equivalence of norms.

Question

Suppose we have two norms on a vector space such that a linear functional is continuous with respect to one if and only if it is continuous with respect to the other. Show that the two norms are equivalent. Two norms $||.||_1 $ and $||.||_2$ are equivalent if there are some constants $C_1,C_2$ such that $C_1||.||_1≤||.||_2≤C_2||.||_1$

Hints, please.

Let's start at the basics, what is the definition of equivalent norms? — Jürgen Sukumaran, May 31 '18 at 14:07
It has already been asked very recently: Non equivalent norms. — mechanodroid, May 31 '18 at 14:53
@TonyS.F. $||.||_1$ and $||.||_2$ are equivalent if there are some constants $C_1$,$C_2$ such that $C_1||.||_1\leq ||.||_2\leq C_2 ||.||_1$ — MathCosmo, May 31 '18 at 15:12
@MathCosmo Why do you think that your question is different...? — saz, Jun 03 '18 at 17:51
@MathCosmo The linked question basically asks the contrapositive of your question: given two nonequivalent norms on a vector space, does there exists a linear functional continuous in one norm and discontinuous in the other? I have clarified it in an answer. — mechanodroid, Jun 03 '18 at 18:03
@MathCosmo improve your question and add this definitions to your question. — miracle173, Jun 03 '18 at 18:06

score 2 · Accepted Answer · answered Jun 03 '18 at 18:02

An answer to the linked question shows that if $\|\cdot\|_1$ and $\|\cdot\|_2$ are two norms on a vector space $X$ such that for any linear functional $f : X \to \mathbb{F}$ holds

$$f \text{ continuous w.r.t. } \|\cdot\|_2 \implies f \text{ continuous w.r.t. } \|\cdot\|_1$$

then there exists $M > 0$ such that $\|\cdot\|_2 \le M\|\cdot\|_1$.

Now, your assumption is that for any linear functional $f : X \to \mathbb{F}$ holds

$$f \text{ continuous w.r.t. } \|\cdot\|_2 \iff f \text{ continuous w.r.t. } \|\cdot\|_1$$

The above statement used in both directions gives that there exist constants $m, M > 0$ such that $\|\cdot\|_2 \le M\|\cdot\|_1$ and $\|\cdot\|_1 \le m\|\cdot\|_2$.

Rearranging gives

$$\frac1m\|\cdot\|_1 \le \|\cdot\|_2 \le M\|\cdot\|_1$$

so $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent norms on $X$.

okay, now I understand this...sorry I didn't take a careful look in that link. — MathCosmo, Jun 04 '18 at 03:28

Basic functional analysis question: equivalence of norms.

1 Answers1