Let $\mathcal{H}$ be a real Hilbert space with norm $\lVert\cdot\rVert$, $\mathcal{B}:\mathcal{H}\times\mathcal{H}\to\mathbb{R}$ be symmetric, bilinear, bounded, coerciv, $\mathcal{L}:\mathcal{H}\to\mathbb{R}$ be linear, bounded. Then define$$\mathcal{I}(u)=\dfrac{1}{2}\mathcal{B}(u,u)-\mathcal{L}(u).$$The problem is
For every $n\in\mathbb{N}$, the variational problem$$\mathcal{I}(u)=\min_{v\in V_n}\mathcal{I}(v)$$has a unique solution $u_n$, where $\{V_n\}_{n\in\mathbb{N}}\subset\mathcal{H}$ is a sequence of closed linear subspace that exhausts $\mathcal{H}$. Moreover, $\{u_n\}_{n\in\mathbb{N}}$ converges to the solution of the variational problem$$\mathcal{I}(u)=\min_{v\in\mathcal{H}}\mathcal{I}(v).$$
My thoughts is using the property of $\{V_n\}$, for $u_m\in V_m$ and any $\delta>0$ we can find a $v_n\in V_n$ such that $\lVert u_m-v_n\rVert\leq\delta$, then I try to use the triangle inequality to prove $\{u_n\}$ is a Cauchy sequence but failed. Are there any better approaches to include it?
Any suggestion will be appreciated.
Observe that $\mathcal{B}(\cdot,\cdot)$ defines an equivalent inner product on $\mathcal{H}$ with induced norm $||u||_{\mathcal{B}}=\sqrt{\mathcal{B}(u,u)}$. Hence, the Riesz representation theorem gives us a $f \in \mathcal{H}$ such that $\mathcal{L}(v)=\mathcal{B}(f,v)$ for all $v \in \mathcal{H}$. Up to a constant factor of $\frac{1}{2}\mathcal{B}(f,f)$, the quantity to be minimized is equal to $$\frac{1}{2}\mathcal{B}(u,u)-\mathcal{L}(u)+\frac{1}{2}\mathcal{B}(f,f) = \frac{1}{2}\mathcal{B}(u-f,u-f)=\frac{1}{2}||u-f||_{\mathcal{B}}^2.$$ If we minimizes this over $\mathcal{H}$, then the unique minimizer is $f$, whereas if we minimize over $V_n$, the solution $u_n$ will be the (unique) best approximation of $f$ in $V_n$ w.r.t. $||\cdot||_{\mathcal{B}}$. Although we do not need it, a well-known fact from Hilbert space theory states that this best approximation is the orthogonal projection onto $V_n$, where orthogonality is understood with respect to $\mathcal{B}$.
Finally, to show the convergence $u_n \to f$, that is, the convergence of the minimizers, we take $\delta>0$ and $N$ such that $\min_{u \in V_n}||u-f|| < \delta$ for all $n \geq N$. This is possible since the sets $V_n$ exhaust $\mathcal{H}$. Now, we compute for $n \geq N$ that $$||u_n-f|| \leq C||u_n-f||_{\mathcal{B}} = C\min_{u \in V_n} ||u-f||_{\mathcal{B}} \leq C' \min_{u \in V_n} ||u-f|| \leq C' \delta,$$ where we have used the boundedness and coercivity property of $\mathcal{B}$, and the fact that $u_n$ minimizes $||\cdot -f||_{\mathcal{B}}$ over $V_n$. The convergence now follows.