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I think I might be completely off on the following question, so would appreciate your opinions and advice:

Given: $$ T:PC[0,1]\to PC[0,1]$$ $$ Tf(x)=\frac{1}{\sqrt{2}}f\big(\frac{x}{2}\big) $$

(where $PC$ stands for piecewise-continuous functions) I have shown that $T$ is bounded and can therefore be extended to: $$ \hat{T}:L_2[0,1]\to L_2[0,1]$$

I am now trying to find $\hat{T}$'s kernel and image.

For the kernel, I figured that it would be sufficient to identify the kernel for $T$ (more precisely, since $T$ is bounded and therefore continuous we get, by definition of the expansion: $\hat{T}f:=\lim Tf_n=T\lim f_n = Tf$ ,for $f_n \to f$ so $Tf=0$ iff $\hat{T}f=0$).

Doing this I get:

$$ Tf=0 \iff \frac{1}{\sqrt2}f\big(\frac{x}{2}\big)=0, \forall x \in [0,1] \iff f(y)=0 ,\forall y \in [0,\frac{1}{2}]$$

and therefore $KerT = g\chi_{(\frac{1}{2},1]}$ for all $g \in PC[0,1]$.

For some reason, I'm uncertain about this.

Also, I do not know how to reach a similar 'closed-form' definition for $ImgT$ - how do I proceed from the definition:

$ImgT = \{Tf(x)=\frac{1}{\sqrt{2}}f\big(\frac{x}{2}\big) \mid f \in PC[0,1]\}$?

I thought of trying to characterise it by plugging in $x$ values to get a feel for what's in the image, but that doesn't feel like the right way to go.

Thanks.

Anon
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  • I don't know what is PC but it seems to be a space of function. Anyway, if PC denotes s subspace of continuous functions, the kernel of $T$ will not be the one you specified for $g\mathbf{1}_{(1/2,1]}$ may fail to be continuous. It will rather be the set of continuous functions in PC with support in $(1/2, 1].$ If PC is measurable, then you are good to go. – William M. May 10 '22 at 18:48
  • If you want to be precise, show that $T$ respects the equivalence relationship on $L^2$, namely if $f=g$ almost everywhere, $Tf = Tg$ almost everywhere. Then to inspect the kernel of $T$ take $f$ in the kernel, a representant, do the computations, conclude. For the image I am pretty sure $T(C^0) = C^0$ and then extend this by density. Remark: for $PC$ to be a subset of $L^2$ you should say precisely what is $PC$, for instance $1/x \notin L^2(0,1)$. – blamethelag May 10 '22 at 19:36
  • @blamethelag Thanks. So with regards to the image, if I understand correctly, I should get that it is all of $L^2$, right? – Anon May 10 '22 at 20:48
  • @Anon yes, also after thoughts I think everything is trivial is you come back to the definition of your map. For sure what you started will work but you can do as follows. For $functions$ define $T$, that is $T : \mathcal L ^2(0,1) \longrightarrow \mathcal L ^2(0,1)$, show it is compatible with the almost everywhere identification, by the quotient properties you have a map $\tilde{T} : L^2 \longrightarrow L^2$ such that denoting $\pi$ the quotient map $\tilde{T} \circ \pi = \pi \circ T$. Then you are left to study $T$ and everything is trivial. – blamethelag May 10 '22 at 21:04

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The same formula defining $T$ on $PC[0,1]$ may be used to define a bounded operator on $L^2[0,1]$. This operator clearly coincides with $T$ on $PC[0,1]$, so it is $\hat T$.

The kernel of $\hat T$ is therefore the subset of $L^2[0,1]$ formed by all functions vanishing on $[0,\frac 12]$, while the image is all of $L^2[0,1]$ because any $f$ in $L^2[0,1]$ is $T(g)$ for $$ g(x) =\left\{\matrix{ \sqrt 2 f(2x), & \text{ if }x<1/2\cr 0, &\text{ otherwise}. }\right. $$

Ruy
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