# Spectrum and compactness of $L^1$ operator

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?

• 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. Jan 4, 2021 at 18:15
• 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. Jan 4, 2021 at 18:17
• Surely something like ${h(t) \over g(t)-\lambda}$ should feature above? Jan 4, 2021 at 18:20
• @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.
– user860263
Jan 4, 2021 at 18:27
• @copper.hat Yes, for $\lambda \neq 0$ $f$ is equal to your expression. In my partial attempt I only considered $\lambda = 0$.
– user860263
Jan 4, 2021 at 18:28

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 $$\sigma(T)=\{0\}$$.

• 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. Jan 5, 2021 at 3:19
• I am not sure why $\sigma_p(T) = \{ \lambda : m(\{ f = \lambda \} ) \}$ is the point spectrum. Shouldn't it be $m(\{ g = \lambda \})$?
– user860263
Jan 5, 2021 at 15:38
• It should be $g$, it's a typo. Jan 5, 2021 at 17:09