"Let $X=C[0,1]$ and $v \in X$ be a fixed function. Let $T$ be the multiplication operator by $v$, i.e. $Tx(t)=v(t)x(t)$. Find the spectrum of $T$."

This is an exercise from a PDF of notes I found online. I'm trying to better understand Spectral Theory.

So $\lambda$ is a regular value if $(T-\lambda I)^{-1}$ exists, is bounded, and dense (I think there is a lemma which lets us not worry about the dense part). The set of all regular values is $\rho(T)$ and the spectrum is $\sigma(T)=\mathbb{C}\setminus \rho(T)$.

It seems to me that $(T-\lambda I)^{-1}$ maps some $y(t)$ to $\frac{y(t) + \lambda}{v(t)}$. However, this would be problematic if $v(t)=0$ for some values of of $t \in [0,1]$

I don't have much of an understanding of all these definitions so if someone could give a solution to this example, I think that would help clear up some of the ideas for me.

Thanks in advance.

  • 2
    $\begingroup$ For the search engines: This is Kreyszig, Section 7.3 - Question #1. $\endgroup$
    – yoshi
    Feb 19, 2017 at 15:14

3 Answers 3


The map $(T-\lambda I)^{-1}$ is defined as the solution mapping of the equation $$ (T-\lambda I) y = z, $$ i.e. $ y= (T-\lambda I)^{-1}z$. In case of the multiplication with $v$, this is equivalent to $$ (v(t)-\lambda) y(t) = z(t). $$ First consider the case that $\lambda\ne v(t)$ for all $t\in [0,1]$. Then the above equation has a unique solution, moreover, $(T-\lambda I)^{-1}$ can be shown to be bounded.

Second, suppose there is $t_0$ such that $\lambda = v(t_0)$. Then $((T-\lambda I)y)(t_0)=0$ for all $y\in X$, hence the range of $(T-\lambda I)$ and its closure are not equal to $X$. Thus, the spectrum of $T$ is the set of all function values of $v$.


I'll be verbose.

For a bounded operator $T$, the resolvent set $\rho(T)$ is the set of $\lambda$ for which $(T-\lambda I)$ is injective and surjective. That is, $\lambda\in\rho(T)$ iff $(T-\lambda I)x=y$ has a unique solution $x\in C[0,1]$, regardless of the choice of $y\in C[0,1]$.

Claim: $\sigma(T) = \mathcal{R}(v)$ where $\mathcal{R}(v)=\{ v(t) : 0 \le t \le 1\}$

First show $\rho(T) \subseteq \mathbb{C}\setminus\mathcal{R}(v)$. To do this, suppose $\lambda \in \rho(T)$ so that there exists $x \in C[0,1]$ such that $(v-\lambda)x=1$. Clearly $\lambda \notin \mathcal{R}(v)$ because that would contradict $(v-\lambda)x=1$. So $\lambda\in\rho(T) \implies \lambda\in\mathcal{C}\setminus\mathcal{R}(v)$, as stated.

Next show $\mathbb{C}\setminus\mathcal{R}(v) \subseteq\rho(T)$. To do this assume $\lambda\notin\mathcal{R}(v)$. Because $v$ is continuous then $\mathcal{R}(v)$ is compact and, therefore, there exists $\epsilon > 0$ such that $|v(t)-\lambda| \ge \epsilon$ for all $t \in [0,1]$. Hence, $x=y/(v-\lambda) \in C[0,1]$ and $(T-\lambda I)x=y$ has a solution $x \in C[0,1]$ for every $y \in C[0,1]$. To see that such a solution $x$ is unique for $\lambda\in\mathbb{C}\setminus\mathcal{R}(v)$, suppose $(T-\lambda I)x_1=y$ and $(T-\lambda I)x_2=y$. Then $(v-\lambda)(x_1-x_2)=0$; however $v-\lambda \ne 0$ on $[0,1]$ because $\lambda\in\mathbb{C}\setminus\mathcal{R}(v)$, which gives $x_1-x_2=0$ on $[0,1]$.

  • $\begingroup$ @DisintegratingByPartsWhy you observe that $|v(t)-\lambda|\ge\varepsilon$? $\endgroup$
    – Jack J.
    Jun 26, 2020 at 20:16

Variation of T.A.E.'s answer to: $\mathbb{C}\setminus\mathcal{R}(v)\subseteq\rho(T)$

Suppose $\lambda\notin\mathcal{R}(v)$.

Regard $w:=\frac{1}{v-\lambda}$.

Then it exists and it is continuous. Moreover, it is bounded as it lives on a compact set.

Thus, its multiplication is an everywhere welldefined and bounded operator: $$\|M_w g\|_\infty\leq\|w\|_\infty\cdot\|g\|_\infty,\quad g\in\mathcal{C}([0,1])$$

Especially, its multiplication is the inverse: $$M_w(M_v-\lambda)f=\frac{1}{v-\lambda}(v-\lambda)f=f$$ $$(M_v-\lambda)M_wg=(v-\lambda)\frac{1}{v-\lambda}g=g$$

Concluding $\lambda\in\rho(T)$.


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