# Resolvent Mapping is Continuous

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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 $T \in \mathfrak{L}(B, B)$. Let $\rho(T)$ be the resolvent set of $T$ in the complex plane. Then the resolvent mapping $f : \rho(T) \to \mathfrak{L}(B,B)$ given by $f(z) = (T - zI)^{-1}$ is continuous in the operator norm $\|\cdot\|_*$.

## Proof

Pick $z\in\rho(T)$. Since $z\in\rho(T)$, the operator $R_z = (T - zI)^{-1}$ exists and has finite norm $C \geq 0$.

Since Resolvent Set is Open, $z+h\in\rho(T)$ for any $h\in \Bbb C$ smaller than some $\delta > 0$. For such $h$,

\(\ds \norm{ f(z+h) - f(z) }_*\) | \(=\) | \(\ds \norm{ (T-(z+h)I)^{-1} - (T-zI)^{-1} }_*\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm{ ((T-zI)-hI)^{-1} - (T-zI)^{-1} }_*\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm{ [(T-zI)(I-h(T-zI)^{-1})]^{-1} - (T-zI)^{-1} }_*\) | by factoring out $T-zI$ from inside the first term | |||||||||||

\(\ds \) | \(=\) | \(\ds \norm{ (I-h(T-zI)^{-1})^{-1}(T-zI)^{-1} - (T-zI)^{-1} }_*\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm{ \big[(I-h(T-zI)^{-1})^{-1} - I\big] (T-zI)^{-1} }_*\) | ||||||||||||

\(\ds \) | \(\leq\) | \(\ds \norm{ (I-h(T-zI)^{-1})^{-1} - I }_* \norm{(T-zI)^{-1} }_*\) | by Operator Norm on Banach Space is Submultiplicative. | |||||||||||

\(\ds \) | \(=\) | \(\ds C\norm{ (I-hR_z)^{-1} - I }_*.\) | (1) |

Consider $(I - hR_z)^{-1}$. Either $\|R_z\|_* = 0$ so that $\|hR_z\|_* = |h|\|R_z\|_* = 0$ (by Operator Norm is Norm) for all $h$, or else $\|hR_z\|_* < 1$ for $h < 1/\|R_z\|_*$. In either case, $\|hR_z\|_* < 1$ for all sufficiently small $h$.

Therefore, by Invertibility of Identity Minus Operator, for sufficiently small $h$ we have $(I - hR_z)^{-1} = I + hR_z + h^2R_z^2 + \ldots$. Substituting this into $(1)$, we get

\(\ds \norm{ f(z+h) - f(z) }_*\) | \(\leq\) | \(\ds C\norm{ hR_z + h^2R_z^2 + \ldots }_*\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds C\norm{hR_z (I + hR_z + h^2R_z^2 + \ldots) }_*\) | ||||||||||||

\(\ds \) | \(\leq\) | \(\ds C\size{h}\norm{R_z}_*\norm{I + hR_z + h^2R_z^2 + \ldots }_*\) | by Operator Norm on Banach Space is Submultiplicative | |||||||||||

\(\ds \) | \(\leq\) | \(\ds C^2\size{h} \sum_{n\in N} (\size{h}\norm{R_z})^n\) | by Triangle Inequality and Operator Norm on Banach Space is Submultiplicative again on each term | |||||||||||

\(\ds \) | \(\leq\) | \(\ds 2C^2\size{h}\) |

as long as $|h| \leq \frac{1}{2\norm{R_z}}$.

The expression on the right-hand side goes to zero as $h$ gets small. Therefore, taking limits, we get

$$ \lim_{h\to0} \norm{ f(z+h) - f(z) }_* = 0. $$

This establishes continuity of $f$ at arbitrary $z\in\rho(T)$.

$\blacksquare$