Let $X= L^2([0,1],\#) $ ( where $\#$ is the measure that counts) $ g\in L^\infty([0,1],\#)$ $A(f)(x)=f(x)g(x)$. Calculate spectrum, eigen values..

Question

Let $X= L^2([0,1],\#) $ ( where $\#$ is the measure that counts) $ g\in L^\infty([0,1],\#)$ and $A \in \mathcal{L} $ as $A(f)(x)=f(x)g(x)$.

i)Show $\sigma(A) = \overline{g([0,1])}$

ii) Determine the eigenvalues and their respective subspaces

iii)Show $A$ is compact iff there exists a sequence {$ a_n$}$_{n \in \mathbf{N} } $ s.t.$g(x)\neq 0$ iff $x = a_n$ for some $n \in \mathbf{N} $ and $ g(a_n) \rightarrow 0$ when $n\rightarrow \infty$

i)To see what values are in the spectrum , I have tryed to do $f(x)g(x)=\lambda f(x)$ so it tells me that the $\lambda = g(x)$ and $\lambda \in \sigma(A) $ and this is the spectrum if I am not wrong, but don't understand exatcly what the closure means .

ii) To determine the eigenvalues I have to take $\lambda $ and see if $A-\lambda I$ is injective so that $\lambda \in \sigma_p(A) $ and their respective subspace would be $ ker(A-\lambda I)$ So I have to determine when $ f(x)g(x)-\lambda f(x)$ is injective . But I am a bit lost here.

iii) This one I don't know how to start it.

Answer 1

Hints:

If $\lambda =g(a)$ for some $a$ and $f(x)=0$ for $x \neq a$, $f(x)=1$ for $x=a$ then $Af=\lambda f$ so $g[0,1] \subset \sigma(A)$. The spectrum of any bounded operator is closed so $\overline {g[0,1]} \subset \sigma(A)$. If $\lambda \notin \overline {g[0,1]}$ the $A-\lambda I$ is invertible with bounded inverse $h \to \frac h {g-\lambda}$. This proves i).

$Af=\lambda f$ with $f \neq 0$ iff $\lambda =g(x)$ for some $c$ and $f(x) =0$ whenever $g(x) \neq \lambda$. This describes all eigen values and eigen vectors.

iii) The spectrum of any compact operator consists of a sequence of numbers tending to $0$ together with $0$ (if it is an infinite set) and this proves one direction of iii). Some hints for the converse: Suppose $(a_n)$ is as in iii). Define $A_nf=fg_n$ where $g_n(x)=0$ if $x \notin \{a_1,a_2,..\}$, $g_n(x_i)=0$ if $i >n$ and $g_n(x_i)=g(x_i)$ for $i \leq n$. Use Heine Borel Theorem to show that each $A_n$ is compact. Then verify that $\|A_n-A\| \to 0$. Since limits in operator norm of compact operators is compact the proof is finished.

Answer 2

Some preliminary observations:

Here we are working with the measure space $([0,1],\mathscr{A},\#)$ where $A\in\mathscr{A}$ iff either $A$ is countable or $[0,1]\setminus A$ is countable. For simplicity, we will refer to $L_p([0,1],\mathscr{A},\#)$ as $L_p$ only.

• Notice that in this space, $\#(A)=0$ iff $A=\emptyset$.

• A function $f\in L_2($ iff $\{f>0\}$ is countable, say $\{x_n:n\in\mathbb{N}\}$ and $\|f\|^2_2=\sum_n|f(x_n)|^2<\infty$. In particular, every function of the form $f_c(x)=\mathbb{1}_{\{c\}}(x)$ is in $L_2$ and $\|f_c\|_2=1$.

• A function $g\in L_\infty$ if there is $m>0$ such that $\{|g|>m\}=\emptyset$. If $\|g\|_\infty:=\inf\{m>0:\{g>m\}=\emptyset\}$, then $\{g>\|g\|_\infty\}=\emptyset$ and so $g\leq\|g\|_\infty$ on $[0,1]$ and $\|g\|_\infty=\sup\{|g(x)|:x\in [0,1]\}$.

• It is a well known fact that if $T$ is a compact operator on a Banach space (for instance $L_2$) the spectrum of $\sigma(T)$ is at most countable and $\sigma(T)\setminus\{0\}$ is made of countable set of of eigenvalues $\{\lambda_n\}$ that accumulate at $\{0\}$ if $\sigma(T)\setminus\{0\}$ is infinite.

• The space of compact operators on Banach space $(X,\|\;\|_X)$ is a closed subspace of $\mathcal{L}(X,X)$ (with the operator norm $\|T\|=\sup_{\|x\|_X=1}\|Tx\|_X$).

• It is a well known fact that the spectrum $\sigma(T)$ of a bounded operator $T\in\mathcal{L}(X,X)$, $X$ a Banach space, is a closed set in $\mathbb{C}$ contained in the ball $B(0;\|T\|)$

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Back to you problem.

(i) It is clear that if $g\in L_\infty$, then $\|fg\|_2\leq\|g\|^2_\infty\|f\|^2_2<\infty$ for all $f\in L_2$ so the operator $A:L_1\rightarrow L_2$ given by $Af=fg$ is indeed bounded and $\|A\|\leq\|g\|_\infty$ (in fact $\|A\|=\|g\|_\infty$).

If $\lambda\in g([0,1])$ and say, $g(c)=\lambda$, then for $f_c=\mathbb{1}_{\{c\}}$ $$ \big(Af_c\big)(x) =f_c(x)g(x)=g(x)\mathbb{1}_{\{c\}}(x)=g(c)f_c(x)=\lambda f_c(x),\quad x\in [0,1]$$ This shows that every point $\lambda\in g([0,1])$ is an eigenvalue of $A$, and hence in the spectrum $\sigma(A)$. Conversely, if $\lambda$ is an eigenvalue of $A$, then for some $f\in L_2$, $f\neq0$, $f(x)g(x)=\lambda f(x)$. In particular, for $x_0$ such that $f(x_0)\neq0$, we have that $g(x_0)=\lambda$ and so, $\lambda\in g([0,1])$. This show that the point spectrum (set of eigenvalues) of $A$ is $g([0,1])$. As a consequence, $\overline{g([0,1])}\subset\sigma(A)$ and so $$\mathbb{C}\setminus\sigma(A)\subset\mathcal{C}\setminus\overline{g([0,1])}\tag{1}\label{one}$$

If $\lambda\in\mathbb{C}\setminus \overline{g([0,1]}=\operatorname{Int}\big(\mathbb{C}\setminus g([0,1])\big)$, then $\alpha:=\inf_{x\in[0,1]}|g(x)-\lambda|>0$ and so, $ h_\lambda=\frac{1}{g-\lambda}\in L_\infty$ with $\|h_\lambda\|_\infty=\frac{1}{\alpha}$. This implies that $fh_\lambda\in L_2$ for all $f\in L_2$, and $$(A-\lambda I)fh_\lambda=(g-\lambda)\frac{f}{g-\lambda}=f$$ Since the operator $B_\lambda:L_2\rightarrow L_2$ given by $f\mapsto h_\lambda f$ is bounded, we have that $(A-\lambda)^{-1}=B_\lambda$ is bounded. Hence, by definition of $\sigma(A)$, $\lambda\in\mathbb{C}\setminus\sigma(A)$ which, together with $\eqref{one}$, means that $\overline{g([0,1])}=\sigma(A)$.

(ii) If $Af=gf$ is compact, then by (1) and the observations above $\sigma(A)$ is (at most) countable and $\sigma(A)\setminus\{0\}\subset g([0,1])$. Necessity of the statement in (ii) follows.

As for sufficiency, suppose $\{|g|>0\}=\{a_n:n\in\mathbb{N}\}\subset[0,1]$ and $g(a_n)\xrightarrow{n\rightarrow\infty}0$. The observations made above imply that $\{g(a_n):n\in\mathbb{N}\}$ are eigenvalues and accumulate at $0$ and so, $\overline{g[0,1]}=\{g(a_n):n\in\mathbb{N}\}\cup\{0\}$. Let $g_n(x)=\mathbb{1}_{\{a_1,\ldots,a_n\}}(x)g(x)$ and $A_nf = g_n f$. Each $A_n$ is a compact operator since it has finte range, that is $\mathbb{R}(A_n)=\operatorname{span}\{\mathbb{1}_{\{a_1\}},\ldots, \mathbb{1}_{\{a_n\}}\}$. Also, $$ \|A_n-A\|\leq\|g_n-g\|_\infty=\sup_{\{m>n\}}|g(a_m)|\xrightarrow\inf_n\sup_{m>n}|g(a_m)|=\limsup_n|g(a_n)|=0 $$ since $g(a_n)\xrightarrow{n\rightarrow\infty}0$. Thus, being a limit (in $\mathcal{L}(L_2,L_2)$ with the operator norm) of a sequence of compact operators, $A$ is itself a compact operator.