# Find the norms of the linear operator $Af(x) =f(x^2)$

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

• Which norm are you using in $C\bigl([0,1]\bigr)$? Mar 22 '18 at 15:54
• @JoséCarlosSantos The max norm Mar 22 '18 at 15:56
• 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. Mar 22 '18 at 15:58
• @Argyll Changed! Thank you Mar 22 '18 at 16:02
• How is A linear? Are you sure you've copied the problem correctly? Mar 22 '18 at 16:04

For a) and b), the answer is $1$. It is clear that $\|Af\|\leqslant\|f\|$ and, if you take $f=1$, you get that $\|Af\|=\|f\|=1$.
• @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, $\|f\|=\|Af\|=1$. Mar 22 '18 at 16:33
• @Acccumulation: not true. $A(2f)=2Af$. Look closer at the definition of $A$ Mar 23 '18 at 22:51