Space of Bounded Linear Transformations forms Vector Space

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 from $X$ to $Y$.

Then $\struct {\map B {X, Y}, +_B, \circ_B}_\GF$ is a vector space.

It therefore suffices to show that $\map B {X, Y}$ is a vector subspace of $\map L {X, Y}$.

From One-Step Vector Subspace Test, it suffices to show that $\map B {X, Y} \ne \O$ and:

$T + \lambda S \in \map B {X, Y}$

for each $\lambda \in \GF$ and $T, S \in \map B {X, Y}$.

First, we have:

$\norm {\mathbf 0_{\map L {X, Y} } x}_Y = 0 \le \norm x_X$

So $\mathbf 0_{\map L {X, Y} } \in \map B {X, Y}$.

Now let $T, S \in \map B {X, Y}$.

Then there exists $M, M' > 0$ such that:

$\norm {T x}_Y \le M \norm x_X$ for each $x \in X$

and:

$\norm {S x}_Y \le M' \norm x_X$ for each $x \in X$.

Then for each $\lambda \in \GF$ and $x \in X$ we have:

$\ds \norm {\paren {T + \lambda S}x}_Y$

$=$

$\ds \norm {T x + \lambda S x}_Y$

$\ds $

$\le$

$\ds \norm {T x}_Y + \norm {\lambda S x}_Y$

$\ds $

$=$

$\ds \norm {T x}_Y + \cmod \lambda \norm {S x}_Y$

Norm Axiom $\text N 2$: Positive Homogeneity

$\ds $

$\le$

$\ds M \norm x_X + \cmod \lambda M' \norm x_X$

So $T + \lambda S \in \map B {X, Y}$.

$\blacksquare$