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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!

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  • $\begingroup$ The operator works as follows: If you take $f \in L^2([0,1])$ and compute $g := Tf$, then we have $g(t) = t \cdot f(t)$ for all $t \in [0,1]$. To show that $T$ has no eigenvectors, assume $f \neq 0$ and $g = Tf = \lambda f$ for some $\lambda \in \Bbb{C}$. Show that this implies $f=0$ (almost everywhere). $\endgroup$
    – PhoemueX
    Commented Jan 26, 2016 at 13:44

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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 $L^2$

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  • $\begingroup$ yes...I will edit it...thanks for pointing this out. $\endgroup$ Commented Jan 26, 2016 at 21:05
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    $\begingroup$ (+1 from me) You might reverse the order of your presentation of the two facts about compactness and eigenvectors. :) $\endgroup$ Commented Jan 26, 2016 at 23:44
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    $\begingroup$ extremely logical suggestion :) $\endgroup$ Commented Jan 27, 2016 at 11:44

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