Why the Galerkin Orthogonality Holds?

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.

• $L$ is given. The problem is to find $u$ or $u_m$ satisfying these variational equations. – daw Feb 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 at 16:15
• because $u$ and $u_m$ solve different problems – daw Feb 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 at 7:43
• The second problem is in a subspace unlike the original one – VorKir Feb 20 at 16:00

$$L$$ 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.