# Spectrum of Bounded Linear Operator is Non-Empty

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## Theorem

Suppose $B$ is a Banach space, $\mathfrak{L}(B, B)$ is the set of bounded linear operators from $B$ to itself, and $T \in \mathfrak{L}(B, B)$. Then the spectrum of $T$ is non-empty.

## Proof

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

We first show that $\|f(z)\|_*$ is uniformly bounded by some constant $C$.

Observe that

- $ \norm{f(z)}_* = \norm{ (T-zI)^{-1} }_* = \frac{1}{|z|} \norm{ (I - T/z)^{-1} }_*. \tag{1}$

For $|z| \geq 2\|T\|_*$, Operator Norm is Norm implies that $\|T/z\|_* \leq \frac{ \|T\|_* }{ 2\|T\|_*} = 1/2$, so by $(1)$ and Invertibility of Identity Minus Operator, we get

\(\ds \norm{f(z)}_*\) | \(=\) | \(\ds \frac{1}{ \size z } \norm{ \sum_{j=0}^\infty \left(\frac{T}{z} \right)^j }_*\) | ||||||||||||

\(\ds \) | \(\leq\) | \(\ds \frac{1}{ \size z } \sum_{j=0}^\infty \frac{\norm T_*^j}{\size z^j}\) | by Triangle Inequality and Operator Norm on Banach Space is Submultiplicative on each term | |||||||||||

\(\ds \) | \(\leq\) | \(\ds \frac{1}{ 2\norm T_* } \sum_{j=0}^\infty \frac{\norm T_*^j}{(2\norm T_*)^j}\) | as $\size z \geq 2\norm T_*$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac{1}{2\norm T_* } \sum_{j=0}^\infty 1/2^j\) | ||||||||||||

\(\ds \) | \(<\) | \(\ds \infty.\) |

Therefore, the norm of $f(z)$ is bounded for $|z| \geq 2\|T\|_*$ by some constant $C_1$.

Next, consider the disk $|z| \leq 2\|T\|_*$ in the complex plane. It is compact. Since $f(z)$ 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 $\|f\|_*$ is bounded on the this disk by some constant $C_2$.

Thus, $\|f(z)\|_*$ is bounded for all $z\in\Bbb C$ by $C = \max \{C_1, C_2\}$.

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

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

\(\ds \size{ g(z) }\) | \(=\) | \(\ds \size{ \ell((T - zI)^{-1} x) }\) | ||||||||||||

\(\ds \) | \(\leq\) | \(\ds \norm{\ell}_{B^*} \norm { (T - zI)^{-1} }_* \norm{x}_B\) | since $\ell$ and $(T-zI)^{-1}$ are bounded by assumption | |||||||||||

\(\ds \) | \(\leq\) | \(\ds \norm{\ell}_{B^*} \norm{x}_B C\) | 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 $|g(z)| \leq \norm{\ell}_{B^*} \norm { (T - zI)^{-1} }_* \norm{x}_B$, together with Resolvent Mapping Converges to 0 at Infinity, implies $|K| = \lim_{z\to\infty} |g(z)| \leq 0$. So $g$ is the constant function $0$.

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

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

$\blacksquare$