Sum of Bounded Linear Transformations is Bounded Linear Transformation

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Theorem

Let $\mathbb F \in \set {\R, \C}$.

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\mathbb F$.

Let $A, B : \HH \to \KK$ be bounded linear transformations.

Let $\norm \cdot$ be the norm on the space of bounded linear transformations.


Then:

$A + B$ is a bounded linear transformation

with:

$\norm {A + B} \le \norm A + \norm B$


Proof

From Addition of Linear Transformations, we have that:

$A + B$ is a linear transformation.

It remains to show that $A + B$ is bounded.

Let $\norm \cdot_\HH$ be the inner product norm on $\HH$.

Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.

Since $A$ is a bounded linear transformation, from Fundamental Property of Norm on Bounded Linear Transformation, we have:

$\norm {A x}_\KK \le \norm A \norm x_\HH$

for all $x \in \HH$.

Similarly, since $B$ is a bounded linear transformation we have:

$\norm {B x}_\KK \le \norm B \norm x_\HH$

for all $x \in \HH$.

Let $x \in \HH$.

Then, we have:

\(\ds \norm {\paren {A + B} x}_\KK\) \(=\) \(\ds \norm {A x + B x}_\KK\)
\(\ds \) \(\le\) \(\ds \norm {A x}_\KK + \norm {B x}_\KK\) Definition of Norm on Vector Space
\(\ds \) \(\le\) \(\ds \norm A \norm x_\HH + \norm B \norm x_\HH\)
\(\ds \) \(=\) \(\ds \paren {\norm A + \norm B} \norm x_\HH\)

So, taking $c = \norm A +\norm B$, we have:

$\norm {\paren {A + B} x}_\KK \le c \norm x_\HH$

for all $x \in \HH$.

So:

$A + B$ is a bounded linear transformation.


Note that:

$\norm A + \norm B \in \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$

while, by the definition of the norm, we have:

$\norm {A + B} = \inf \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$

So, by the definition of infimum:

$\norm {A + B} \le \norm A + \norm B$

$\blacksquare$