Sufficient Condition for Weak Convergence of Bounded Sequence in Hilbert Space in terms of Everywhere Dense Set

Theorem

Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a bounded sequence in $\HH$.

Let $D$ be an everywhere dense set in $\HH$ such that:

$\innerprod {x_n} v \to \innerprod x v$

for all $v \in D$, for some $x \in \HH$.

Then:

$x_n \weakconv x$

where $\weakconv$ denotes weak convergence.

Proof

We show the theorem in the case $x = 0$ first for neatness.

From Weak Convergence in Hilbert Space, we want to show that:

$\innerprod {x_n} z \to \innerprod x z$

for all $z \in \HH$.

Let $\epsilon > 0$.

Since $D$ is everywhere dense in $\HH$, for each $n \in \N$ we can pick $z_n \in D$ such that:

$\norm {z - z_n}_\HH < \dfrac 1 n$

We now have, for each $n \in \N$ and $m \in \N$:

$\cmod {\innerprod {x_n} z} = \cmod {\innerprod {x_n} {z - z_m} - \innerprod {x_n} {z_m} }$

due to the conjugate linearity of the inner product in its second argument.

Pick $M > 0$ such that:

$\norm {x_n}_\HH \le M$

for each $n \in \N$.

We can then bound:

\(\ds \cmod {\innerprod {x_n} {z - z_m}_\HH - \innerprod {x_n} {z_m}_\HH}\)

\(\le\)

\(\ds \cmod {\innerprod {x_n} {z - z_m}_\HH } + \cmod {\innerprod {x_n} {z_m}_\HH}\)

Triangle Inequality for Complex Numbers

\(\ds \)

\(\le\)

\(\ds \norm {x_n}_\HH \norm {z - z_m}_\HH + \cmod {\innerprod {x_n} {z_m}_\HH}\)

Cauchy-Bunyakovsky-Schwarz Inequality for Inner Product Spaces

\(\ds \)

\(\le\)

\(\ds \frac M m + \cmod {\innerprod {x_n} {z_m}_\HH}\)

since $\norm {x_n}_\HH \le M$ and $\norm {z - z_m}_\HH < 1/m$

Fix $m \in \N$ such that:

$m > \dfrac {2 M} \epsilon$

Since we have:

$\innerprod {x_n} {z_m}_\HH \to 0$

for this fixed $m$, we can pick $N$ such that:

$\cmod {\innerprod {x_n} {z_m}_\HH} < \dfrac \epsilon 2$

for $n \ge N$.

Then for $n \ge N$ we have:

$\cmod {\innerprod {x_n} z_\HH} < \epsilon$

So:

$\innerprod {x_n} z_\HH \to 0 = \innerprod 0 z$

giving:

$x_n \weakconv 0 = x$

Now we show the theorem for general $x$.

Suppose that:

$\innerprod {x_n} v \to \innerprod x v$

for all $v \in D$.

Then, using the linearity of the inner product in the first argument, we have:

$\innerprod {x_n - x} v \to 0$

Then by our previous calculations:

$x_n - x \weakconv 0$

so:

$x_n \weakconv x$

from Linear Combination of Weakly Convergent Sequences is Weakly Convergent.

$\blacksquare$