Theorem: Spectrum of Bounded Linear Operator is Non-Empty

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 $f: \C \to \map {\mathfrak L} {B, B}$ be the resolvent mapping defined as $\map f z = \paren {T - z I}^{-1}$.
{{AimForCont}} the spectrum of $T$ is empty, so that $\map f z$ is well-defined for all $z \in \C$. 
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:
{{begin-eqn}}
{{eqn | l = \norm {\map f z}_*
      | r = \frac 1 {\size z} \norm {\sum_{j \mathop = 0}^\infty \paren {\frac T z}^j}_*
}}
{{eqn | o = \le
      | r = \frac 1 {\size z} \sum_{j \mathop = 0}^\infty \frac {\norm T_*^j} {\size z^j}
      | c = by Triangle Inequality and Operator Norm on Banach Space is Submultiplicative on each term
}}
{{eqn | o = \le
      | r = \frac 1 {2 \norm T_*} \sum_{j \mathop = 0}^\infty \frac {\norm T_*^j} {\paren {2 \norm T_*}^j}
      | c = as $\size z \ge 2 \norm T_*$
}}
{{eqn | o \le
      | r = \frac 1 {2 \norm T_*} \sum_{j \mathop = 0}^\infty 1/2^j
}}
{{eqn | o = <
      | r = \infty
}}
{{end-eqn}}
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 $z \in \C$ 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 $g: \C \to \C$ as:
:$\forall z \in \C: \map g z = \map \ell {\map f z x}$
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:
{{begin-eqn}}
{{eqn | l = \size {\map g z}
      | r = \size {\map \ell {\paren {T - z I}^{-1} x} }
}}
{{eqn | o = \le
      | r = \norm \ell_{B^*} \norm {\paren {T - z I}^{-1} }_* \norm x_B
      | c = since $\ell$ and $\paren {T - z I}^{-1}$ are bounded by assumption
}}
{{eqn | r = \norm \ell_{B^*} \norm x_B C
      | o = \le
      | c = by the above
}}
{{eqn | o = <
      | r = \infty
}}
{{end-eqn}}
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 that:
:$\ds \size K = \lim_{z \mathop \to \infty} \size {\map g z} \le 0$
So $g$ is the constant function $0$.
We have therefore shown that:
:$\forall x \in B, \ell \in B^*: \map \ell {\map f z x} = 0$
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.
{{qed}}
Category:Bounded Linear Operators
Category:Spectra (Bounded Linear Operators)
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