Exercise 32 from chapter 4 (“Hilbert Spaces: An Introduction”) of Stein & Shakarchi's “Real Analysis” 2 I have a question about the following problem in Stein and Shakarchi's book.

Consider the operator $T:L^2([0,1]) \to L^2([0,1])$ defined by $$T(f)(t) = tf(t).$$
  (a) Prove that $T$ is a bounded linear operator with $T = T^*$, but that $T$ is not compact.
  (b) However, show that $T$ has no eigenvectors.

First of all, I don't really get how this operator works. Could you guys give me insight into this operator and guide me through these problems? Thanks!
 A: That operator $T$ that is defined above is taking any function $f\in L^2[0,1]$ to the function $tf(t)\in L^2[0,1]$.
$\textbf {Boundedness of T}:$ Note that
$$ \|Tf\|_2^2=\int_{0}^{1}t^2|f(t)|^2dt\leq\int_{0}^{1}|f(t)|^2dt=\|f\|_2^2.$$
Thus, $T$ is a bounded linear operator. Moreover, $\|T\|\leq 1$. (In fact, you can show that $\|T\|=1$.)
$\textbf{T=T*}$: $$\langle Tf,g\rangle =\int_{0}^{1}tf(t)\overline{g(t)}dt= \int_{0}^{1}f(t)\overline{tg(t)}dt = \langle f,Tg\rangle.$$
So, by the uniqueness of the adjoint operator, it follows that $T=T^*$.
$\textbf{T has no eigenvectors}$: First of all recall that an eigenvector is nonzero by definition. Let us assume that $f$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda$. Then
$$ tf(t) = Tf(t) = \lambda f(t),$$
for all $t\in[0,1]$, whence it follows that $f=0$ almost everywhere. This contradiction proves that $T$ has no eigenvector.
$\textbf{T is not compact}$: For this part have a look at the following
Compactness of  Multiplication Operator on $L^2$
