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I have a question. If $u$ and $u_h$ are the solutions of the continuous and discrete variational equations, respectively, then for $u\in H^1_0$, how does one prove that $\lim_{h\rightarrow 0} \|u - u_h\| = 0$ where the norm is taken in the $H^1(\sigma)$. Sigma is the domain. Thank you.

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    Standard homework comment: What have you tried?2012-07-05
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    I'm really not sure as to how to start. I mean, I took the definition of the formulations, took the difference but I can't proceed from there. What properties of the norm do I need to use.2012-07-05
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    I tried to make your notation readable. I could not parse what the 2 was supposed to mean. It looks a little like $u\in H^1_0$?2012-07-05
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    Hello, thank you so much. It should be u ∈, not u 2. Also, the h in uh is a subscript. And the norm is taken in H^1("symbol for sigma here in the bracket, not in the subscript"). I'm really grateful, for I don't know how to type with those notations..2012-07-05
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    Let me know if this is correct, and take a look at the edits to see how it's done :)2012-07-05

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This is a broadly posed question, so I'll paint a picture in similarly broad strokes.

Let $E$ denote the functional you are minimizing on some space $V$. Let $E_*=\inf_V E$, and assume it's attained at $u_*$. You need two ingredients:

(*) $\qquad \forall \epsilon>0$ $\exists \delta>0$ such that $E(u)

(**)$\qquad$The union of all finite-dimensional subspaces $V_h\subset V$ (which correspond to discretization of the problem) is dense in $V$.

Given $\epsilon>0$, pick $\delta>0$ from (*). Since the set $\Omega_\delta=\{u\in V\colon E(v) is open, there exists $h>0$ such that $V_h\cap\Omega_\delta$ is nonempty. The minimum of $E$ on $V_h$ will be attained by some $u_h\in\Omega_\delta$, which guarantees $\|u_h-u_*\|<\epsilon$.