Fundamental Property of Norm on Bounded Linear Transformation
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Theorem
Let $\HH, \KK$ be Hilbert spaces.
Let $A: \HH \to \KK$ be a bounded linear transformation.
Let $\norm A$ denote the norm of $A$ defined by:
- $\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$
Then:
- $\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$
Proof
From Norm on Bounded Linear Transformation is Finite:
- $\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$ exists
and
- $\norm A < \infty$
Let $x \in \HH \setminus \set{0_\HH}$
Let $\lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$.
Then:
\(\ds \norm {A x}_K\) | \(\le\) | \(\ds \lambda \norm x_\HH\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \dfrac {\norm {A x}_\KK} {\norm x_\HH}\) | \(\le\) | \(\ds \lambda\) |
As $c$ was arbitrary, then:
- $\forall \lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}: \dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \lambda$
By the definition of the infimum:
- $\dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \norm A$
Hence:
- $\norm {A x}_\KK \le \norm A \norm x_\HH$
Since $x$ was arbitrary:
- $\forall h \in \HH \setminus \set {0_\HH}: \norm {A h}_\KK \le \norm A \norm h_\HH$
Lastly, we have:
\(\ds \norm {A 0_\HH}_\KK\) | \(=\) | \(\ds \norm {0_\KK}_\KK\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm A \cdot 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm A \norm {0_\HH}\) |
It follows that:
- $\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$
$\blacksquare$
Sources
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- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $\S \text {II}.1$