- Define the spectrum $σ(T)$ of $T$ and the spectral radius $r(T)$ of $T$.
- State without proof a sufficient criterion for the invertibility of $I-T$ and use it to show that $r(T) ⩽\|T\|$. Briefly indicate how this implies the stronger inequality that $r(T) ⩽ \inf_{j ⩾ 1}\|T^j\|^{1 / j}$.

- The spectrum $σ(T)$ consists of complex numbers $λ$ for which $λ I-T$ is not invertible as a bounded linear operator. The spectral radius $r(T)$ of $T$ is defined by $r(T)=\sup_{λ ∈ σ(T)}|λ|$.
- Statement: $I-T$ is invertible if $\|T\|<1$.

Now, if $λ>\|T\|$, then the above implies that $I-\frac{1}{λ} T$, hence $λ I-T$ is invertible, i.e. $λ ∉ σ(T)$. This means $r(T) ⩽\|T\|$.

For the stronger inequality, we use the fact that: If $λ ∈ σ(T)$, then $λ^j ∈σ(T^j)$. This follows that $r(T)^j ⩽\|T^j\|$, whence the desired inequality.

- Quoting without proof a suitable theorem, explain why there exists $ℓ_{x_0} ∈ X^*$ with $\|ℓ_{x_0}\|=1$ such that \[ ℓ_{x_0}(f)=f(x_0) \text { for all } f ∈ C([0,1]) . \]
- It is a fact, which you may use, that the function $ψ_{x_0}(x)=\exp (\frac{\mathrm{i}}{x-x_0})$ satisfies \[ \|g+ψ_{x_0}\|_{L^{∞}((0,1))}-|g(x_0)| ⩾ 1 \text { for all } g ∈ C([0,1]) . \] By considering possible values for $ℓ_{x_0}(ψ_{x_0})$ and the behaviour of $ℓ_{x_0}$ on $C([0,1]) ⊕⟨ψ_{x_0}⟩$, explain why $ℓ_{x_0}$ is not unique.

- Let $ℓ: C([0,1]) → ℂ$ be given by $ℓ(f)=f(x_0)$. Clearly $ℓ$ is bounded linear on $C([0,1])$ with norm $\|ℓ\|=1$. By the Hahn-Banach theorem, there exists an extension of $ℓ$ to $X$, denoted by $ℓ_{x_0}$, satisfying $\|ℓ_{x_0}\|=\|ℓ\|=1$ as desired.
- Note that $ψ_{x_0} ∈ X ∖C([0,1])$. By the Hahn-Banach theorem, we only need to show that the extension of $ℓ$ to $C([0,1]) ⊕⟨ψ_{x_0}⟩=: Y$ is non-unique. To this end, we only need to show that there is more than one way of specifying the value of $ℓ_{x_0}(ψ_{x_0})=:α$. In order that $ℓ_{x_0}$ is bounded on $Y$ with norm $\|ℓ_{x_0}\|_{Y^*}=1$, we need $|g(x_0)+λ α|=\left|ℓ_{x_0}(g+λ ψ_{x_0})\right| ⩽\|g+λ ψ_{x_0}\|_{L^{∞}((0,1))}$ for all $λ ∈ ℂ$ and $g ∈ C([0,1])$.

Rescaling $g$, this is equivalent to \[ |g(x_0)+α| ⩽\|g+ψ_{x_0}\|_{L^{∞}((0,1))} \text { for all } g ∈ C([0,1]) . \] By the triangle inequality, this holds true if $α$ can be selected such that \[ |α| ⩽\|g+ψ_{x_0}\|_{L^{∞}((0,1))}-|g(x_0)| \text { for all } g ∈ C([0,1]) . \] By the property of $ψ_{x_0}$, the above is satisfies for all $|α| ⩽ 1$. This shows that the choices of $α$ and hence of $ℓ_{x_0}$ are non-unique.

- Describe the image of $T$ and find a polynomial $p$ such that $p(T)=0$.
- Show that $\|T\|=\||e_0|+|e_1|\|$.
- Compute $σ(T)$ and verify explicitly that Gelfand's formula holds: $r(T)=\inf_{j ⩾ 1}\|T^j\|^{1 / j}$. [You may use any results on spectra of bounded linear operators on Banach spaces.]

- We have $\operatorname{Im} T=⟨e_0, e_1⟩$ and $T^2=T$, i.e. $p(x)=x^2-x$.
- Recall that $\|ℓ_0\|=\|ℓ_1\|=1$, which implies $|ℓ_0(f)| ⩽\|f\|$ and $|ℓ_1(f)| ⩽\|f\|$. Hence
\[
|T f(x)| ⩽|ℓ_0(f)||e_0(x)|+|ℓ_1(f)||e_1(x)| ⩽\|f\|(|e_0(x)|+|e_1(x)|) ⩽\|f\|\||e_0|+|e_1|\| .
\]
This implies $\|T\| ⩽\||e_0|+|e_1|\|$.

For the converse, let $x_1 ∈[0,1]$ be the point where the continuous function $|e_0|+|e_1|$ attains maximum. Let $f$ be the linear function such that $f(0) e_0(x_1)=∣ e_0(x_1) ∣$ and $f(1) e_1(x_1)=|e_1(x_1)|$. Note that at least $f(0)$ or $f(1)$ is non-zero as $|e_0|+|e_1|$ is non-zero. Hence $\|f\|=1$. It is readily seen that$$Tf(x_1)=|e_0(x_1)|+|e_1(x_1)|=\||e_0|+|e_1|\|$$which implies that $\|T\| ⩾\||e_0|+|e_1|\|$. The conclusion follows. - Since $p(T)=0$, we have by the spectral mapping theorem that $λ^2-λ=0$ for all $λ ∈ σ(T)$, hence $σ(T) ⊂\{0,1\}$. Since $T$ is not surjective, $0 ∈ σ(T)$. Since $T e_0=e_0$, we have $λ=1$ is an eigenvalue of $T$ and so $1 ∈ σ(T)$. Therefore $σ(T)=\{0,1\}$.

Clearly, $r(T)=1$ by definition. Note also that $T^j=T$ for all $j ⩾ 1$. By (ii), $\|T\|=\||e_0|+|e_1|\| ⩾|e_0(0)|+|e_1(0)|=1$ and so $\|T^j\|^{1 / j}=\|T\|^{1 / j}$ is non-increasing and limits to 1. This implies $\inf _{j ⩾ 1}\|T^j\|^{1 / j}=1$ and Gelfand's formula is verified.