Define a linear operator $T:X \to X$ as $ Tf(x) = x^2f(x) $.Prove that its spectrum set is $[0,1]$

Question

Let $X = C([0,1])$ with the supremum norm. Define a linear operator $T:X \to X$ as $$ Tf(x) = x^2f(x) $$ for every $f \in X$ and $x \in [0,1]$. Prove that its spectrum set is $[0,1]$.

I tried to apply the definitions of three kinds of spectrum, but it doesn't go well at all, and I was kind of confused how to apply them correctly.

Answer 1

The spectrum is the set of scalars such that $T - \lambda I$ is not invertible (i.e. either not surjective or not injective). In particular, we have $$(T - \lambda I)(f) = \Big(x \mapsto x^2 f(x) - \lambda f(x) = (x^2 - \lambda)f(x)\Big).$$ Importantly, note that when $\lambda \in [0, 1]$, then $x^2 - \lambda$ has a root in the domain $[0, 1]$. This implies that $T - \lambda I$ is not surjective (why?), and thus, $[0, 1]$ is in the spectrum of $T$.

If $\lambda \notin [0, 1]$, then $x^2 - \lambda$ has no root in $[0, 1]$, and hence $x \mapsto \frac{1}{x^2 - \lambda}$ is a continuous function on $[0, 1]$. See if you can use this to find an inverse of $T$.

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So, you'll hopefully have seen that, if $\lambda \in [0, 1]$, then $x^2 - \lambda = 0$ when $x = \sqrt{\lambda}$. This means that, for every $g$ in the range of $T - \lambda I$, we have $g(\sqrt{\lambda}) = 0$.

So, putting this another way, we can define a linear function $\phi_\lambda : C[0, 1] \to \Bbb{R}$ by $$\phi_\lambda(f) = f(\sqrt{\lambda}).$$ Then, we have $$\operatorname{Range}(T - \lambda I) \subseteq \operatorname{Ker} \phi_\lambda.$$

Since $C[0, 1]$ is equipped with the supremum norm, the map $\phi_\lambda$ is bounded, since $$|\phi_\lambda(f)| = |f(\sqrt{\lambda})| \le \sup_{x \in [0, 1]} |f(x)| = \|f\|.$$ This implies that $\phi_\lambda$ is continuous, and hence its kernel (i.e. $\phi_\lambda^{-1}\{0\}$) is a closed set. (This is not true with the integral norms.)

So, if $\operatorname{Range}(T - \lambda I)$ were dense, then we would have $$C[0, 1] = \overline{\operatorname{Range}(T - \lambda I)} \subseteq \overline{\operatorname{Ker} \phi_\lambda} = \operatorname{Ker} \phi_\lambda \subseteq C[0, 1],$$ and hence $$\operatorname{Ker} \phi_\lambda = C[0, 1].$$ But this is not true! Simply choose $f$ to be, say, the constant function $1$, and $\phi_{\lambda}(f) = 1 \neq 0$. Thus, $\operatorname{Range}(T - \lambda I)$ is not dense.

You can also do the same argument in terms of adjoints. If the range of $T - \lambda I$ were dense, then we would expect $(T - \lambda I)^*$ to be injective. But, $(T - \lambda I)^*(\phi_\lambda) = \phi_\lambda \circ (T - \lambda)$ is the zero functional, hence $\phi_\lambda \in \operatorname{Ker}(T - \lambda I)^*$, proving $(T - \lambda I)^*$ is not injective.