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This is not homework. I'm going over my lecture notes to study for an exam.

For an Abstract Elliptic Problem such as the problem with V a Hilbert Space

$$\begin{cases} \text{Find } u \in V \text{ such that} \\ a(u,v) = L(v) && \forall v\in V \end{cases}$$

where $a$ is a bounded, coercive bilinear form. $L \in V^*$. We know that by Lax-Milgram that $u$ is the unique solution

We search for approximations to $u$, $u_m \in V_m \subset V$ where $V_m$ is a finite dimensional subset of $V$.

In the Galerkin method we seek the best approximation $u_m$ of $u$ with respect to the bilinear form $a$ described above.

$$\begin{cases} \text{Find } u_m \in V_m \text{ such that} \\ a(u_m,v_m) = L(v_m) && \forall v_m\in V_m \end{cases}$$

The Galerkin orthogonality states that:

$$a(u-u_m, v_m) = 0$$

which implies that orthogonality of the error with respect to $a$. The usual proof is given as follows:

$$a(u-u_m, v_m) = a(u,v_m) - a(u_m,v_m) = L(v_m) - L(v_m) = 0$$

What I don't understand is why the two $L$s in the above are the same. Since from Riesz Representation Theorem, $L$ is the unique functional in the dual of $V$ or $V_m$, how can they possibly be equal otherwise $u=u_m$?

Edit: From comments below, the two functionals (L) are not technically operating in the same spaces. On the other hand, they both represent some inner product ($L^2$ or other) and they both produce the same result, hence equality in result without necessarily equivalance.

Not a chance
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    $L$ is given. The problem is to find $u$ or $u_m$ satisfying these variational equations. – daw Feb 19 '19 at 15:45
  • but if $L$ is given, since $a$ is also given how can $u_m$ be different to $u$ since the mapping to the dual space is an isometry? – Not a chance Feb 19 '19 at 16:15
  • because $u$ and $u_m$ solve different problems – daw Feb 19 '19 at 20:37
  • OK, but if the functional analysis theory holds that there is a unique $L$ such that $a(u,v)=L(v) \forall v$ then $a(u_m,\cdot)\neq a(u,\cdot)$ unless $u_m=u$ – Not a chance Feb 20 '19 at 07:43
  • The second problem is in a subspace unlike the original one – VorKir Feb 20 '19 at 16:00
  • Hi thanks for the comment. So the standard proofs are an abuse of notation or something? The two Ls are not the same? – Not a chance Feb 21 '19 at 11:31
  • @daw i guess you voted my question down. Care to tell me why? – Not a chance Feb 21 '19 at 11:39
  • Why not study instead the real thing? At hand of a simple example to begin with: Understanding Galerkin method of weighted residuals . – Han de Bruijn Feb 21 '19 at 15:02
  • yes ok i will add an edit to my question. It is clear to me from VorKir that indeed the integrals are the same, since "f", the source term, will be the same. So the answer is zero. But indeed the two bilinear cannot give the same functional L (as the wikipedia article states since they either refer to the same place or imply that the mapping is not bijective. – Not a chance Feb 23 '19 at 21:37
  • Or should I delete the question? It was downvoted... – Not a chance Feb 23 '19 at 21:39
  • I really don't understand the question. In the second case you use the restriction of $L$ onto the subspace $V_m$. Of course you could call this restriction $L_m$ and write $L_m(v_m)$ instead of $L(v_m)$, but since $L_m(v_m)=L(v_m)$, there is really no difference. – MaoWao Mar 01 '19 at 12:55

1 Answers1

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$L$ is a given linear and continuous function $L:V\to\mathbb R$.

As you write in the comments, there exists a unique functional $M\in V^\star$ such that $ a(u,v) = M(v) \forall v\in V$. (I am giving this object a name different from $L$ to avoid confusion).

Also, there exists a unique functional $M_m\in V_m^\star$ such that $ a(u_m,v_m) = M_m(v_m) \forall v_m\in V_m$.

In general, $M_m\neq M$.

However, the proof uses the functional $L$ and not $M,M_m$, so $L(v_m)-L(v_m)=0$ in the proof is always true.

supinf
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