2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Let's start at the basics, what is the definition of equivalent norms? $\endgroup$ – Tony S.F. May 31 '18 at 14:07
  • 1
    $\begingroup$ It has already been asked very recently: Non equivalent norms. $\endgroup$ – mechanodroid May 31 '18 at 14:53
  • 1
    $\begingroup$ @MathCosmo Why do you think that your question is different...? $\endgroup$ – saz Jun 3 '18 at 17:51
  • 1
    $\begingroup$ @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. $\endgroup$ – mechanodroid Jun 3 '18 at 18:03
  • 1
    $\begingroup$ @MathCosmo improve your question and add this definitions to your question. $\endgroup$ – miracle173 Jun 3 '18 at 18:06
1
+50
$\begingroup$

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 $X$.

$\endgroup$
  • $\begingroup$ okay, now I understand this...sorry I didn't take a careful look in that link. $\endgroup$ – MathCosmo Jun 4 '18 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.