Why does from $\|Tf_n\| \rightarrow 0$ follow that $\lambda \in [0,1]$ are approximate eigenvalues?

Question

I'm studying for my exam (which is actually today) of functional analysis. I'm struggling with the following question:

Define the linear operator $T:L^2(0,1) \rightarrow L^2(0,1)$ by $(Tf)(x)=xf(x)$. Determine the spectrum of T.

The solution we were given is the following:

I don't understand the first part of the solution. I don't understand why for $\|Tf_n\|_2 \rightarrow 0$ follows that $\lambda \in [0,1]$ is an approximate eigenvalue of $T$. I guess that it follows from the following theorem:

Let H be a Hilbert space, and let $T: H \rightarrow H$ be bounded and normal. Then, $\lambda \in \sigma(T)$ iff there is a sequence $(x_n)$ such that $\|x_n\|=1$ for all n, and $(T-\lambda)x_n \rightarrow 0$.

However, we haven't proved that $(T-\lambda)f_n \rightarrow 0$ right? Only that $\|Tf_n\|_2 \rightarrow 0$. But why does from this follow that $[0,1] \subset \sigma(T)$? Furthermore, where does the sequence come from? Why did they take $x\in [\lambda-\frac{1}{n},\lambda+\frac{1}{n}]$?

An explanation would be very much appreciated (since I have my exam today) :) !!

Answer 1

There is a typo in the first part. Actually, $\|(T-\lambda)f_n\|^{2}=2n\int_{\lambda-\frac 1 n}^{\lambda+\frac 1 n} (x-\lambda)^{2}dx =\frac 4 {3n^{2}} \to 0$.

Answer 2

Here is another way to show that $\sigma (T)= [0,1]$. If $\lambda \notin [0,1]$ then for every $f \in L^2(0,1)$, the function $g=\tfrac{1}{\lambda-x} f \in L^2$ and $$ (λ-T)g=f. $$ This shows that $\lambda \in ρ(T) = \mathbb C \setminus \sigma(T)$. Conversely, if $\lambda \in [0,1]$ then for $f(x)=1$ there is no $g \in L^2(0,1)$ such that $(λ-T)g=f$. This is because the function $ x \mapsto \tfrac{1}{λ-x}$ is not in $L^2(0,1)$. This shows that $λ-T$ is not injective and so $\lambda \in \sigma(T)$.

In general, if $u $ is a fixed continuous function and $M_u \colon L^2 \to L^2$ is the multiplication operator given by $(M_u(f))(x) = u(x)f(x)$ then $\sigma(M_u) = Range(u)$. The proof is essentially the same. Moreover, the same result should also be true if $u $ is only assumed to be in $L^2$, where in that case $\sigma(M_u) = ess~range(u)$.