Properties of the linear operator $T(f)=\chi_{[0,\frac{1}{2}]}f(2x)$ where $T:L^1[0,1] \to L^1[0,1]$ Properties of the linear operator $T(f)=\chi_{[0,\frac{1}{2}]}f(2x)$ where  $T:L^1[0,1] \to L^1[0,1]$
I am supposed to find the operator norm, check whether it is one to one or onto and if it is compact.
I think that the norm is $1/2$ as $\int_{0}^{1/2}f(2x)=1/2\int_{0}^{1}f(x)$. It is $1-1$ as $T(f)=0$ iff $f(x)=0$ for all $x$. It is clearly not surjective. I also think that it cannot be compact. I think that the range is closed and not finite dimensional and so $T$ is not compact. IS that true?It is closed beceasue passing into subsequence we get a function that is $0$ on $[1/2,1]$
 A: Thanks for the clarification on the definition. I think that you have figured out the whole problem by yourself!
I agree that the norm is $\frac{1}{2}$, and this follows precisely from the computation you wrote.
I also agree that it can't be compact because the range is indeed closed and not finite dimensional. For the sake of having a complete answer I'll say why those two things are true:

*

*It's clear that the range is
$$
T(L^1([0,1])):= \{ g \in L^1([0,1]): g(x)=0 \ \forall \ x \in (\textstyle{\frac{1}{2}},1] \}
$$
From here we see that the range is infinite dimensional, in fact it's isomorphic (not isometrically) to $L^1([0,\frac{1}{2}])$.

*Finally to see that the range is closed we take any sequence $(f_n)$ in $T(L^1([0,1]))$ such that $f_n \to f$ in $L^1([0,1])$. The main goal is to show that $f \in T(L^1([0,1]))$. Well, as you pointed out there is a subsequence of $(f_n)$ that converges $a.e$ to $f$ and from this it follows that $f(x)=0$ for a.e. $x \in [\frac{1}{2},1]$. Let $\tilde{f}$ be the function gotten from $f$ by letting $\tilde{f}(x)=0$ for all $x \in [\frac{1}{2},1]$ for which $f(x) \neq 0$. Then $f=\tilde{f}$ a.e. on $L^{1}([0,1])$ (so they are the same element) and there fore $f_n \to \tilde{f} \in T(L^1([0,1]))$.

