Compact Linear Transformation is Bounded

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Theorem

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

Let $\map {B_0} {\HH, \KK}$ be the space of compact linear transformations $\HH \to \KK$.

Let $\map B {\HH, \KK}$ be the space of bounded linear transformations $\HH \to \KK$.

Let $T \in \map {B_0} {\HH, \KK}$.


Then $T \in \map B {\HH, \KK}$.

That is:

$\map {B_0} {\HH, \KK} \subseteq \map B {\HH, \KK}$


Proof

Let $T : \HH \to \KK$ be a linear transformation.

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

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

We show that:

if $T$ is not bounded, then it is not compact.

That is:

if $T \not \in \map B {\HH, \KK}$, then $T \not \in \map {B_0} {\HH, \KK}$.

Then, we will obtain:

$\map {B_0} {\HH, \KK} \subseteq \map B {\HH, \KK}$


Suppose that $T$ is not bounded.

That is:

there does not exist a real number $c > 0$ such that $\norm {T h}_\KK \le c \norm h_\HH$ for all $h \in \HH$.

Then:

for each $n \in \N$, there exists some $x_n \in \HH$ such that $\norm {T x_n}_\KK > n \norm {x_n}_\HH$.

Now define sequence $\sequence {h_n}$ by:

$\ds h_n = \frac {x_n} {\norm {x_n} }$

for each $n \in \N$.

We have:

$\norm {h_n} = 1$

for each $n \in \N$, so $\sequence {h_n}$ is bounded.

We also have:

$\norm {T h_n} > n$

for each $n \in \N$.

So, for any increasing sequence of positive integers $\sequence {n_j}$ we have:

$\norm {T h_{n_j} } > n_j$

Note that $n_j \to \infty$.

So, for every subsequence $\sequence {h_{n_j} }$ of $\sequence {h_n}$ we have that:

$\sequence {T h_{n_j} }$ is unbounded.

So from Convergent Sequence in Normed Vector Space is Bounded, we have that:

$\sequence {T h_{n_j} }$ does not converge.

So:

the bounded sequence $\sequence {h_n}$ does not have a subsequence $\sequence {h_{n_j} }$ such that $\sequence {T h_{n_j} }$ converges.

So:

$T$ is not compact.

Hence the demand.

$\blacksquare$


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