# Point spectrum of a particular operator

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 $$[a,b]\subseteq [0,1]$$, $$a such that $$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 $$\lambda\in \sigma_p(T)$$.

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 $$a, i.e. $$[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\}.$$