Spectrum of Bounded Linear Operator is Non-Empty

Theorem
Let $B$ be a Banach space.

Let $\map {\mathfrak L} {B, B}$ be the set of bounded linear operators from $B$ to itself.

Let $T \in \map {\mathfrak L} {B, B}$.

Then the spectrum $\map \sigma T$ of $T$ is non-empty.

Proof
Let $f : \Bbb C \to \map {\mathfrak L} {B, B}$ be the resolvent mapping defined as $\map f z = \paren {T - z I}^{-1}$. Suppose the spectrum of $T$ is empty, so that $\map f z$ is well-defined for all $z \in \Bbb C$.

We first show that $\norm {\map f z}_*$ is uniformly bounded by some constant $C$.

Observe that:


 * $(1): \quad \norm {\map f z}_* = \norm {\paren {T - z I}^{-1} }_* = \frac 1 {\size z} \norm {\paren {I - T/z}^{-1} }_*$

For $\size z \ge 2 \norm T_*$, Operator Norm is Norm implies that $\norm {T / z}_* \le \dfrac {\norm T_*} {2 \norm T_*} = 1/2$, so by $(1)$ and Invertibility of Identity Minus Operator, we get

Therefore, the norm of $\map f z$ is bounded for $\size z \ge 2 \norm T_*$ by some constant $C_1$.

Next, consider the disk $\size z \le 2 \norm T_*$ in the complex plane. It is compact. Since $\map f z$ is continuous on the disk by Resolvent Mapping is Continuous, and since Norm is Continuous, we get from Continuous Function on Compact Space is Bounded that $\norm f_*$ is bounded on the this disk by some constant $C_2$.

Thus, $\norm {\map f z}_*$ is bounded for all $z \in \Bbb C$ by $C = \max \set {C_1, C_2}$.

Finally, pick any $x \in B$ and $\ell \in B^*$, the dual of $B$. Define the function $g : \Bbb C \to \Bbb C$ by $\map g z = \map \ell {\map f z x}$.

Since $f$ has empty spectrum, Resolvent Mapping is Analytic and Strongly Analytic iff Weakly Analytic together imply that $g$ is an entire function. Thus we have

So $g$ is a bounded entire function. It is therefore equal to some constant $K$ by Liouville's Theorem.

But the inequality above $|\map g z| \le \norm \ell_{B^*} \norm {\paren {T - z I}^{-1} }_* \norm x_B$, together with Resolvent Mapping Converges to 0 at Infinity, implies $\ds \size K = \lim_{z \mathop \to \infty} \size {\map g z} \le 0$. So $g$ is the constant function $0$.

We have therefore shown that $\map \ell {\map f z x} = 0$ for any $x \in B, \ell \in B^*$. This implies from Condition for Bounded Linear Operator to be Zero that $\map f z = 0$, and in particular that $\map f 0 = T^{-1} = 0$.

But this is a contradiction, since our assumption that the spectrum of $T$ is empty implies that $T$ has a two-sided bounded inverse.