Norm on Space of Bounded Linear Transformations is Norm

From ProofWiki
Jump to navigation Jump to search

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:

$\norm {T + S}_{\map B {X, Y} } \le \norm T_{\map B {X, Y} } + \norm S_{\map B {X, Y} }$

$\blacksquare$

Retrieved from "https://proofwiki.org/w/index.php?title=Norm_on_Space_of_Bounded_Linear_Transformations_is_Norm&oldid=632984"