Consider a continuous complex function $g$ defined on $[0,1]$. On $L^1([0,1])$ consider operator $$Tf(x) = g(x)f(x).$$ Calculate the norm. Determine its point spectrum and spectrum. Decide for what functions $g$ is $T$ compact.

I've managed to show that $\lVert T \rVert = \lVert g \rVert_\infty$.

To find point spectrum we want to find nontrivial solutions $$ \forall x \in [0,1]: (\lambda - g(x)) f(x) = 0.$$

For constant function $g$ we get a nontrivial solution. From this $\sigma_p(T) = \{g = c \in \mathbb{C} \}$.

For nonconstant $g$ we have no nontrivial solutions. Which means $\sigma_p(T) = \emptyset$.

To find spectrum we choose $h \in L^1([0,1])$ and we need to find for what $\lambda \in \mathbb{C}$ the following equation has a solution $$ \forall x \in [0,1]: \lambda f(x) - g(x) f(x) = h(x). $$

If we set $\lambda = 0$, then $f(x) = -h(x)/g(x)$ for $g \neq 0$ a. e. Since I distinct constant and noncostant functions, we get that for constant $g$ we always have a solution. Which means $0 \notin \sigma(T) $. Since a compact operator always has $0$ in its spectrum, for such funtions $g$ operator $T$ cannot be compact.

For $g$ nonconstant we need to decide if $-h(x)/g(x) \in L^1$. That is where I am unsure about how to proceed.

Is my partial attempt correct? How would I go about the compactness?

  • $\begingroup$ Mostly correct, but for the spectrum, you need to consider a general $\lambda$, not just $\lambda=0$. Note also that if $h\in L^1$ and $g\in L^\infty$ then $h/g\in L^1$ if $g$ is never zero. $\endgroup$ Jan 4, 2021 at 18:15
  • $\begingroup$ Don't need $g$ to be globally constant; If $g$ is constant on a set of nonzero measure, then $T$ has eigenvectors consisting of all functions supported on that set. $\endgroup$ Jan 4, 2021 at 18:17
  • $\begingroup$ Surely something like ${h(t) \over g(t)-\lambda}$ should feature above? $\endgroup$
    – copper.hat
    Jan 4, 2021 at 18:20
  • $\begingroup$ @Chrystomath thanks for your suggestions. I know that I need to check for a general $\lambda$. I started with $0$, since sometimes it is easier. $\endgroup$
    – user860263
    Jan 4, 2021 at 18:27
  • $\begingroup$ @copper.hat Yes, for $\lambda \neq 0$ $f$ is equal to your expression. In my partial attempt I only considered $\lambda = 0$. $\endgroup$
    – user860263
    Jan 4, 2021 at 18:28

1 Answer 1


Let us look at the eigenvalues first. You what $g(x)f(x)=\lambda\,f(x)$ a.e. Let $E=\{f\ne0\}$. For any $x\in E$, you have $g(x)=\lambda$; and the equality holds trivially on the complement of $E$. So $\lambda$ is an eigenvalue of $g$ if and only if $m(E)>0$. Conversely, if $m(\{g=\lambda\}>0$, we have $g\,1_E=\lambda 1_E$ and $\lambda$ is an eigenvalue. Thus $$ \sigma_p(T)=\{\lambda:\ m(\{g=\lambda\})>0\}. $$

For the whole spectrum, you need to ask yourself when $g-\lambda$ is invertible. It is easy to see that if $1/(g-\lambda)$ is continuous, it provides an inverse $T-\lambda I$. Because $g$ is continuous and $[0,1]$ is compact, it is enough that $g(x)\ne\lambda$ for all $x$. In other words: if $\lambda$ is not in the range of $g$, then $\lambda\not\in\sigma(g)$. That is, $\sigma(T)\subset g([0,1])$.

And conversely, suppose $g(x)=\lambda$ for some $x$. As $g$ is uniformly continuous, for each $n$ there exists $\delta_n$ such that $|x-y|<\delta_n$ implies $|g(x)-g(y)|<1/n$. Let $f_n=\frac1{\delta_n}\,1_{[x-\delta_n/2,x+\delta_n/2]}$. Then $f_n\in L^1[0,1]$, $\|f_n\|_1=1$. And
$$ \|Tf_n-\lambda f_n\|_1=\frac1{\delta_n}\,\int_{[x-\delta_n/2,x+\delta_n/2]}|g(t)-f(x)| \,dt\leq\frac1n. $$ So $T-\lambda$ cannot be invertible, and so $\lambda=g(x)\in\sigma(T)$, showing that $g([0,1])\subset \sigma(T)$. Thus $$\tag1 \sigma(T)=g([0,1]). $$

Finally, compactness. As you noted, for $T$ to be compact a necessary condition is that $0\in\sigma(T)$, which means that $g(x)=0$ for some $x$. More generally a necessary condition for compactness is that the spectrum consists of $0$ and a sequence of eigenvalues that converge to zero. In light of $(1)$, this requires $g$ to take at most countably many values. By the Intermediate Value Theorem, this can only happen if $g$ is constant; but this would require $g$ to be zero. So $T$ can only be compact when $g=0$.

An even more telling way to look at compactness is to notice that, if $\lambda$ is an eigenvalue for $T$, then $\dim\ker(T-\lambda I)=\infty$. Because if $\lambda$ is an eigenvalue then $g(x)=\lambda$ on some measurable set $E$ with $m(E)>0$. If we write $E$ as a disjoint union $E=E_1\cup\cdots\cup E_m$, then $1_{E_1},\ldots,1_{E_m}$ are $m$ linearly independent eigenvectors for $\lambda$; and we can do this for any $m$. This is not a proof, though, because it may happen that $\sigma(T)=\{0\}$.

  • $\begingroup$ Another very informal way of looking at compactness (with $g$ continuous) is that the range 'contains' $L^1(E)$ in some sense and so cannot be compact. $\endgroup$
    – copper.hat
    Jan 5, 2021 at 3:19
  • $\begingroup$ I am not sure why $\sigma_p(T) = \{ \lambda : m(\{ f = \lambda \} ) \}$ is the point spectrum. Shouldn't it be $m(\{ g = \lambda \})$? $\endgroup$
    – user860263
    Jan 5, 2021 at 15:38
  • $\begingroup$ It should be $g$, it's a typo. $\endgroup$ Jan 5, 2021 at 17:09

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