1
$\begingroup$

Find the 1-norms of the linear operator $Af(x) =f(x^2)$ if:

a) $A: C[0,1] \rightarrow C[0,1]$

b) $A: C[-1,1] \rightarrow C[0,1]$

c) $A: L^1(0,1) \rightarrow L^2(0,1)$

I honestly have no idea how to do this. For part a, my guess is: $\|Af\|= \|f(x^2)\| \leq \|f(x)\| \forall f \in C[0,1] $? then I don't know what to do from here.

For part b, I think you get the same inequality except here its $\forall f \in C[-1,1]$, but again I don't know how to conclude this.

For part c, we have $\|Af(x)\|_1 \leq \|f(x^2)\|_1$

I keep seeing examples online where they just let be equal to some value such as $f \equiv 1 $ in order to show $\|A\| \geq$ whatever bound we get from above, but I don't understand why that's ok or if I can do the same here.

EDIT: I understand parts a and b, but I'm still confused on part c

$\endgroup$
  • 1
    $\begingroup$ Which norm are you using in $C\bigl([0,1]\bigr)$? $\endgroup$ – José Carlos Santos Mar 22 '18 at 15:54
  • $\begingroup$ @JoséCarlosSantos The max norm $\endgroup$ – Vinny Chase Mar 22 '18 at 15:56
  • $\begingroup$ you should add 1- after "Find the ". I would edit it for you. But stackexchange somehow thinks edits on math questions need to have minimal character length. $\endgroup$ – Argyll Mar 22 '18 at 15:58
  • $\begingroup$ @Argyll Changed! Thank you $\endgroup$ – Vinny Chase Mar 22 '18 at 16:02
  • $\begingroup$ How is A linear? Are you sure you've copied the problem correctly? $\endgroup$ – Acccumulation Mar 22 '18 at 16:04
1
$\begingroup$

For a) and b), the answer is $1$. It is clear that $\|Af\|\leqslant\|f\|$ and, if you take $f=1$, you get that $\|Af\|=\|f\|=1$.

$\endgroup$
  • $\begingroup$ But why can you just take f=1? $\endgroup$ – Vinny Chase Mar 22 '18 at 16:02
  • $\begingroup$ because that's one of the only two norm-1 constant functions? $\endgroup$ – Argyll Mar 22 '18 at 16:07
  • $\begingroup$ @VinnyChase I proved that $\|A\|\geqslant1$. So, if I find a function $f$ such that $\|f\|=\|Af\|=1$, that will prove that $\|A\|=1$. And it turns out that if you take $f=1$, then, indeed, $\|f\|=\|Af\|=1$. $\endgroup$ – José Carlos Santos Mar 22 '18 at 16:33
  • $\begingroup$ Thank you. What about for c? $\endgroup$ – Vinny Chase Mar 22 '18 at 20:08
  • 1
    $\begingroup$ @Acccumulation: not true. $A(2f)=2Af$. Look closer at the definition of $A$ $\endgroup$ – Argyll Mar 23 '18 at 22:51

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.