# Spectrum of Bounded Linear Operator is Non-Empty

## Theorem

Let $B$ be a Banach space over $\C$.

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 $\map f z = \paren {T - z I}^{-1}$.

Aiming for a contradiction, 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:

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

For $\size z \ge 2 \norm T_*$, Operator Norm is Norm implies that $\norm {\dfrac T z}_* \le \dfrac {\norm T_*} {2 \norm T_*} = \dfrac 1 2$.

Hence by $(1)$ and Invertibility of Identity Minus Operator, we get:

 $\ds \norm {\map f z}_*$ $=$ $\ds \frac 1 {\size z} \norm {\sum_{j \mathop = 0}^\infty \paren {\frac T z}^j}_*$ $\ds$ $\le$ $\ds \frac 1 {\size z} \sum_{j \mathop = 0}^\infty \frac {\norm T_*^j} {\size z^j}$ by Triangle Inequality and Operator Norm on Banach Space is Submultiplicative on each term $\ds$ $\le$ $\ds \frac 1 {2 \norm T_*} \sum_{j \mathop = 0}^\infty \frac {\norm T_*^j} {\paren {2 \norm T_*}^j}$ as $\size z \ge 2 \norm T_*$ $\ds$ $=$ $\ds \frac 1 {2 \norm T_*} \sum_{j \mathop = 0}^\infty 1/2^j$ $\ds$ $<$ $\ds \infty$

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.

From Resolvent Mapping is Continuous $\map f z$ is continuous on the disk.

From 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:

 $\ds \size {\map g z}$ $=$ $\ds \size {\map \ell {\paren {T - z I}^{-1} x} }$ $\ds$ $\le$ $\ds \norm \ell_{B^*} \norm {\paren {T - z I}^{-1} }_* \norm x_B$ since $\ell$ and $\paren {T - z I}^{-1}$ are bounded by assumption $\ds$ $\le$ $\ds \norm \ell_{B^*} \norm x_B C$ by the above $\ds$ $<$ $\ds \infty$

So $g$ is a bounded entire function.

It is therefore equal to some constant $K$ by Liouville's Theorem.

But the inequality above $\size {\map g z} \le \norm \ell_{B^*} \norm {\paren {T - z I}^{-1} }_* \norm x_B$, 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 $\map f z = 0$.

In particular:

$\map f 0 = T^{-1} = 0$

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

$\blacksquare$