Let $X=C[0,1]$ and we define

$$T\colon X\to X\quad\text{as}\quad(Tf)(t)=g(t)f(t)\quad\text{for all}\quad t\in [0,1]$$ with $g\in C[0,1]$ fixed. I have proved that $\sigma(T)=g([0,1])$.

Question 1. If I prove that $\lVert T \rVert =\lVert g \rVert_{\infty}$ I can coclude that the operator $T$ is bounded?

Now, I must find the point spectrum. This is my attempt:

If $\lambda\in \sigma_p (T)$, then exists an $f\in X\setminus\{0\}$ such that $Tf=\lambda f$. Since $f\ne 0$ in continuous there exists an interval $[a,b]\subseteq [0,1]$, $a<b$ such that $f(t)\ne 0$ on $[a,b]$. But this implies that for all $t\in [a,b]$ we have $$g(t)f(t)=\lambda f(t)\iff g(t)=\lambda.$$

Therefore $\lambda\in M_g$, where $$M_g:=\big\{\lambda\in C\;|\;\text{exists}\; 0\le a< b\le 1\;\text{such that}\; m|_{[0,1]}=\lambda\big\}.$$

Vice versa, if $\lambda\in M_g$, then we have an interval $[a,b]\subseteq [0,1]$ such that $(T-\lambda I)f=0$ on $[a,b]$ and for all $f\in X$.

Question 2. Can we find a function $f\ne 0$ with $f(t)=0$ for all $t\notin [a,b]$? If yes, why?

In fact if this function exists, then $(T-\lambda I)=0$ on $[0,1]$, i,e $\lambda\in \sigma_p(T)$.


1 Answer 1


For question 1: If you prove that $\|T\|$ is finite then you can conclude that $T$ is bounded so yes, of course if $\|T\|=\|g\|_\infty$, $T$ is bounded. Actually, $\|T(f)\|_\infty=\|g\cdot f\|_\infty\leq\|g\|_\infty\|f\|_\infty$ for any $f\in C[0,1]$, so $\|T\|\leq\|g\|_\infty<\infty$ and this is enough to conclude that $T$ is bounded.

For the point spectrum: if $\lambda\in\sigma_p(T)$, then there exists a continuous function $f\in C[0,1]$ that is not zero everywhere such that $g(t)f(t)=\lambda f(t)$ for all $t\in[0,1]$. If $S=\{x\in[0,1]: f(x)\neq0\}$, then $S$ is an open subset of $[0,1]$ and it is true that $g(t)=\lambda$ for all $x\in S$. Since $S$ is not empty and open, there does exist an interval $[a,b]\subset S$, thus for all $t\in[a,b]$ it is $g(t)=\lambda$. Conversely: suppose that $\lambda\in\mathbb{C}$ has the property "there exists $[a,b]\subset[0,1]$ such that $g(t)=\lambda$ for all $t\in[a,b]$. Choose numbers $c,d$ such that $a<c<d<b$, i.e. $[c,d]$ is a proper subset of $[a,b]$. Define a function $f(t)$ on $[0,1]$ such that $f$ is equal to $1$ on $[c,d]$, $f$ is equal to $0$ on $[0,a]$ and $[b,1]$. Draw a graph and extend $f$ linearly in $[a,c]$ and $[d,b]$. It is immediate that $g(t)f(t)=\lambda f(t)$. This function is $0$ outside of $[a,b]$, as you want.

Conclusion: $$\sigma_p(T)=\{\lambda\in\mathbb{C}: \text{ there exists }[a,b]\subset[0,1]\text{ such that for all }t\in[a,b]: g(t)=\lambda\}.$$


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