Sequence of subspaces is dense. Limit of Operator norm Let $V \subset L^2(\Omega)$ be a Hilbertspace and $\{V_n\}$ a sequence of subspaces such that
\begin{align*}
V_1 \subset V_2 \subset \dots \quad \text{and} \quad \overline{\bigcup_{n \in \mathbb{N}} V_n} = V \, (\text{w.r.t. } V\text{-norm} ).
\end{align*}
For some $f\in L^2(\Omega)$ we define $\phi_n = \sup_{\| v_n\| = 1, v_n \in V_n} \int_\Omega f(x) v_n(x)\, dx$. How can I prove that
\begin{align*}
\lim_{n\to\infty} \phi_n = \sup_{\| v\| = 1, v \in V} \int_\Omega f(x) v(x)\, dx
\end{align*} 
holds? Is this convergence uniform?
 A: No that convergence is not uniform, because it has nothing to be uniform over. $\phi_n$ is not a function, it is just a real number, depending only on $n$. 
To show your result, prove:


*

*$\Phi := \sup_{\| v\| = 1, v \in V} \int_\Omega f(x) v(x)\, dx < \infty$.

*$\phi_n$ is an increasing sequence, bounded above by $\Phi$.

*For any value less than $\Phi$, there must be some $\phi_n$ greater than that value.


You can conclude that $\phi_n \to \Phi$ from that (do you see why?)
A: Clearly $\phi_n \le  \phi = \sup_{\| v\| = 1, v \in V} \int_\Omega f(x) v(x)\, dx, \forall n \in \mathbb{N}$.
Let $\varepsilon > 0$. We wish to prove that $\exists n \in \mathbb{N}$ such that  $\phi - \phi_n < \varepsilon$.
Set $\delta = \min\left\{\frac{\varepsilon}{4\|f\|_2}, \frac{\varepsilon}4, \frac12\right\} > 0$ and pick $v \in V$, $\|v\| = 1$ such that $\phi - \int_\Omega fv < \delta$.
Since $\overline{\bigcup_{n\in\mathbb{N}}V_n} = V$, there exists $n \in \mathbb{N}$ and $v_n \in V_n$ such that $\|v-v_n\|_2 < \frac\delta2$. Set $v_n' = \frac{v_n}{\|v_n\|}$. We have $\|v_n'\|_2 = 1$  and
$$\|v_n - v_n'\|_2 = \left\|v_n\left(1-\frac1{\|v_n\|}\right)\right\|_2 = \left|1-\|v_n\|_2\right| = \left|\|v\|-\|v_n\|_2\right| \le \|v-v_n\| < \frac\delta2$$
so $\|v-v_n'\|_2 \le \|v - v_n\|_2 + \|v_n - v_n'\|_2 < \delta$.
Now 
\begin{align}
\left|\phi - \phi_n\right| &\le \left|\phi - \int_\Omega fv\right| + \left|\int_\Omega fv - \int_\Omega fv_n'\right| + \left|\int_\Omega fv_n' - \int_\Omega fv_n\right| + \left|\int_\Omega fv_n - \phi_n\right| \\
&< \delta + \|f\|_2\|v-v_n'\|_2 + \|f\|_2\|v_n'-v_n\|_2 + \delta\\
&\le  \frac{\varepsilon}4 + \|f\|_2\frac{\varepsilon}{4\|f\|_2} + \|f\|_2\frac{\varepsilon}{4\|f\|_2} +\frac{\varepsilon}4\\
&= \varepsilon
\end{align}
Since $\varepsilon$ was arbitrary,  we conclude $\lim_{n\to\infty} \phi_n = \phi$.
