Question:
Let $H=L^2([0,1],m)$, where $m$ is the Lebesgue measure, and consider th e operators $M,S\in L(H,H)$ given by
$$ Mf(t)=tf(t), \; Sf(t)=f(1-t), f\in H, t\in [0,1]$$
You are not asked to verify that $M$ and $S$ defined bounded linear operators on $H$.\
Justify that $M$ is self-adjoint, but not compact.
Show that $\lVert Sf\rVert_2 = \lVert f \rVert_2$, for all $f\in H$. Does $S$ have any eigenvalues? Is $S$ compact?
My Attempt:
Note for $f_1\in H$
$$\langle M_t f,f_1\rangle=\int_{[0,1]} tf(t)\overline{f_1(t)}dm(t)=\int_{[0,1]} f(t)\overline{tf_1(t)}dm(t)=\langle f_1, M_{t}f\rangle$$
where we used $t\in [0,1]\subset \mathbb{R}$. Thus it is self-adjoint. Moreoever by (something we proved in class) it follows that $M$ has no eigenvalues, yet $\sigma(M)=[0,1]$, and so suppose for the sake of contradiction that $M$ is indeed compact. Then each non-zero $\lambda \in [0,1]$ would also be an eigenvalue of $M$, contradictory to the fact that $M$ has no eigen-values.
Note
$$ \langle Sf,f_1\rangle=\int_{[0,1]}f(1-t)\overline{f_1(t)}dm(t)=\int_{[0,1]}f(u)\overline{f(1-u)}dm(u)=\langle f,Sf\rangle$$
0 Where we simply made a change of variables. To see $\lVert Sf\rVert_2=\lVert f \rVert_2$ note;
$$\lVert Sf\rVert=\int_[0,1] f(1-t)^2 dm(t)=\int_[0,1] f(u)^2dm(u)=\lVert f\rVert
$$
where we made a simple change of variables. So we have shown $S$ is an isometry. Since we have isometry it follows that
$$\langle f,f \rangle=\langle Sf, Sf\rangle=\langle \lambda f, \lambda f\rangle =\vert \lambda\vert^2 \langle f,f\rangle$$
from which we clearly see the only possible eigenvalues are $-1,1$. Suppose for the sake of contradiction that $S$ is compact. Then we must have $\sigma(S)\setminus{\{0\}}=EV(S)\setminus{\{0\}}$. But $\sigma(S)\setminus{\{0\}}\neq \{-1,1\}=EV(S)\setminus{\{0\}}$ and hence we have reached a contradiction and we conclude that $S$ is not compact.
My questions:
Does the proof look right?
Best Answer
What you say about $M$ is fine. As soon as you know that $\sigma(M)=[0,1]$ you know it cannot be compact, because a compact operator only admits $0$ as an accumulation point of the spectrum.
What you say about $S$ is not right, in the sense that you seem to be claiming that $\sigma(S)\ne\{-1,1\}$. But $\sigma(S)=\{-1,1\}$, so your argument does not apply. In fact, both $1$ and $-1$ are eigenvalues of $S$ (for instance with eigenvectors $f(t)=|t-\frac12|$ and $f(t)=1-2t$ respectively), so the nonzero part of the spectrum of $S$ does agree with the set of eigenvalues.
What you can say about $S$ is that since $S^2=I$ you have that $S$ is invertible, and hence it cannot be compact.