Norm on Space of Bounded Linear Transformations is Norm
Theorem
Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$.
Let $\map B {X, Y}$ be the space of bounded linear transformations between $X$ and $Y$.
Let $\norm {\, \cdot \,}_{\map B {X, Y} }$ be the norm on the space of bounded linear transformations.
Then $\norm {\, \cdot \,}_{\map B {X, Y} }$ is indeed a norm on $\map B {X, Y}$.
Proof
From Norm on Bounded Linear Transformation is Finite, $\norm {\, \cdot \,}_{\map B {X, Y} }$ is real-valued.
Proof of Norm Axiom $\text N 1$: Positive Definiteness
First, we have:
- $\norm {\mathbf 0_{\map B {X, Y} } x}_Y = 0$
for each $x \in X$.
So:
- $\set {\norm {\mathbf 0_{\map B {X, Y} } x}_Y : \norm x_X = 1} = \set 0$
That is:
- $\norm {\mathbf 0_{\map B {X, Y} } }_{\map B {X, Y} } = 0$
Now suppose that $T \in \map B {X, Y}$ has $\norm T_{\map B {X, Y} } = 0$.
Then from Fundamental Property of Norm on Bounded Linear Transformation, we have:
- $\norm {T x}_Y \le \norm T_{\map B {X, Y} } \norm x_X = 0$
for each $x \in X$.
Then from Norm Axiom $\text N 1$: Positive Definiteness for $\norm {\, \cdot \,}_Y$, we have $T x = 0$ for each $x \in X$.
So $T = \mathbf 0_{\map B {X, Y} }$.
This proves Norm Axiom $\text N 1$: Positive Definiteness.
$\Box$
Proof of Norm Axiom $\text N 2$: Positive Homogeneity
Take $\lambda \in \GF$ with $\lambda \ne 0$, $T \in \map B {X, Y}$ and $x \in X$.
Take $M > 0$ such that:
- $\norm {T x}_Y \le M \norm x_X$
Then:
\(\ds \norm {\lambda T x}_Y\) | \(=\) | \(\ds \cmod \lambda \norm {T x}_Y\) | Norm Axiom $\text N 2$: Positive Homogeneity for $\norm {\, \cdot \,}_Y$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds M \cmod \lambda \norm x_X\) | by hypothesis |
Taking the infimum over $M > 0$ we have:
- $\norm {\lambda T x}_Y \le \cmod \lambda \norm T_{\map B {X, Y} } \norm x_X$
So:
- $\norm {\lambda T}_{\map B {X, Y} } \le \cmod \lambda \norm T_{\map B {X, Y} }$
Now suppose that:
- $\norm {\lambda T}_{\map B {X, Y} } < \cmod \lambda \norm T_{\map B {X, Y} }$
Then there exists $m < \cmod \lambda \norm T_{\map B {X, Y} }$ such that:
- $\norm {\lambda T x}_Y \le m \norm x_X$
for each $x \in X$.
Then by Norm Axiom $\text N 2$: Positive Homogeneity for ${\, \norm \,}_Y$ we have:
- $\ds \norm {T x}_Y \le \frac m {\cmod \lambda} \norm x_X$
where:
- $\ds \frac m {\cmod \lambda} < \norm T_{\map B {X, Y} }$
This contradicts the definition of $\norm T_{\map B {X, Y} }$, so we must have:
- $\norm {\lambda T}_{\map B {X, Y} } = \cmod \lambda \norm T_{\map B {X, Y} }$
$\Box$
Proof of Norm Axiom $\text N 3$: Triangle Inequality
Let $T, S \in \map B {X, Y}$.
Then, for each $x \in X$ we have:
\(\ds \norm {\paren {T + S} x}_Y\) | \(=\) | \(\ds \norm {T x + S x}_Y\) | Definition of Linear Transformation | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {T x}_Y + \norm {S x}_Y\) | Norm Axiom $\text N 3$: Triangle Inequality for $\norm {\, \cdot \,}_Y$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \paren {\norm T_{\map B {X, Y} } + \norm S_{\map B {X, Y} } } \norm x_X\) | Fundamental Property of Norm on Bounded Linear Transformation |
Taking the supremum over $\norm x_X = 1$, we have:
- $\norm {T + S}_{\map B {X, Y} } \le \norm T_{\map B {X, Y} } + \norm S_{\map B {X, Y} }$
$\blacksquare$