Problem
Let's observe in interval $[a,b]\subset \mathbb{R}$ defined real valued and continuous maps which form vector space $\mathcal{F}([a,b],\mathbb{R})$. Let $\langle ., . \rangle$ be some inner product.
Now let's observe equation
$$ Lf=u $$
where $L$ is a linear transformation in space $\mathcal{F}$ and $f,u \in \mathcal{F}$. Let $v \in \mathcal{F}$ be arbitrary. Now previous equations finite-dimensional weak form is
$$ \langle Lf,v \rangle = \langle u , v \rangle \quad \forall v \in \mathcal{F} .$$
Let $V_d \subset \mathcal{F}$ be subspace where dim$(V_d)=n \in \mathbb{N}$. First equations solution $f$ projection $f_d$ to subspace $V_d$ satisfies finite dimensional weak form:
$$ \langle Lf_d,v_d \rangle = \langle u , v_d \rangle \quad \forall v_d \in V_d .$$
Now we can call the following as error vector of projection
$$ (f-f_d) \in \mathcal{F}. $$
Show that image of error vector $L(f-f_d)$ belongs to the orthogonal complement of subspace $V_d$.
Attempt to show
By definition, something belongs to the orthogonal complement of $V_d$ when the following condition is satisfied. Let's denote orthogonal complement of $V_d$ as $V_d^{\perp}$
$$ V_d^{\perp} = \{ \vec{v} \in \mathcal{F} | \langle \vec{v}, \vec{w} \rangle = 0 \quad \forall \vec{w} \in V_d \} $$.
So I need to somehow show that $L(f-f_d) \in V_d^{\perp}$.
I've been trying to think about this but it seems I'm having little to no progress towards a solution. Any hints or suggestions on what should I do? I think I'm stuck.
Edit
I think I need to show that (note I can be wrong)
$$ L(f-f_d) \in V_d \iff \langle L(f-f_d),w \rangle = 0 \quad \forall w \in V_d $$
$$ \iff \langle Lf,w \rangle - \langle Lf_d, w \rangle = 0 \quad \forall w \in V_d $$
$$ \iff \langle Lf,w \rangle = \langle Lf_d, w \rangle \quad \forall w \in V_d $$
Now the equation is begin to look like finite dimensional weak form
If we utilize the fact that $Lf=u$ we can rewrite the equation as
$$ \iff \langle Lf_d , w \rangle = \langle u, w \rangle \quad \forall w \in V_d $$
Now when $V_d$ should satisfy this by "definition" this should be it?