Norm of Adjoint
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Theorem
Let $H, K$ be Hilbert spaces.
Let $A \in \map B {H, K}$ be a bounded linear transformation.
Let $A^* \in \map B {K, H}$ be the adjoint of $A$.
Then $A$ and $A^*$ satisfy:
- $\norm A_{\map B {H, K} }^2 = \norm {A^*}_{\map B {K, H} }^2 = \norm {A^* A}_{\map B {H, H} }$
where:
- $\norm \cdot_{\map B {H, K} }$ denotes the operator norm on $\map B {H, K}$
- $\norm \cdot_{\map B {K, H} }$ denotes the operator norm on $\map B {K, H}$
- $\norm \cdot_{\map B {H, H} }$ denotes the operator norm on $\map B {H, H}$
Proof
Let $h \in H$ such that $\norm h_H \le 1$.
Then:
\(\ds \norm {A h}_K^2\) | \(=\) | \(\ds \innerprod {A h} {A h}_K\) | Definition of Inner Product Norm | |||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod {A^* A h} h_H\) | Definition of Adjoint Linear Transformation | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {A^*A h}_H \norm h_H\) | Cauchy-Bunyakovsky-Schwarz Inequality | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {A^* A} \norm h_H^2\) | Fundamental Property of Norm on Bounded Linear Transformation | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {A^* A}\) | Assumption on $h$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {A^*} \norm A\) | Norm on Bounded Linear Transformation is Submultiplicative |
By definition $(1)$ for $\norm A$, it follows that:
- $\norm A^2 \le \norm {A^* A} \le \norm {A^*} \norm A$
That is:
- $\norm A \le \norm {A^*}$.
By substituting $A^*$ for $A$, and using $A^{**} = A$ from Adjoint is Involutive, the reverse inequality is obtained.
Hence $\norm A^2 = \norm {A^* A} = \norm {A^*}^2$.
$\blacksquare$
Sources
- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\text {II}.2.7$
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $13.1$: Existence of Hilbert Adjoint