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
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 for $\norm {\, \cdot \,}_Y$, we have $T x = 0$ for each $x \in X$.

So $T = \mathbf 0_{\map B {X, Y} }$.

This proves.

Proof of
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:

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 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:


 * $\norm {\lambda T}_{\map B {X, Y} } = \cmod \lambda \norm T_{\map B {X, Y} }$

Proof of
Let $T, S \in \map B {X, Y}$.

Then, for each $x \in X$ we have:

Taking the supremum over $\norm x_X = 1$, we have:


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