Problem
Let's observe in a closed interval $[a,b] \subset \mathbb{R}$ real-valued and continuous vectorspace $\mathcal{F}([a,b],\mathbb{R})$. Where $\langle ., .\rangle$ is some scalar product. This scalar product only satisfies axioms when $\overline{v},\overline{w},\overline{u} \in V$ and $c \in \mathbb{R}$ (V is space for scalar product)
- $\langle \overline{v}, \overline{w} \rangle = \langle \overline{w}, \overline{v} \rangle$
- $\langle \overline{v}, \overline{w} + \overline{u} \rangle = \langle \overline{v}, \overline{w} \rangle + \langle \overline{v}, \overline{u} \rangle$
- $\langle c \overline{v}, \overline{w} \rangle = c \langle \overline{v}, \overline{w} \rangle $
- $\langle \overline{v}, \overline{v} \rangle \ge 0$; and $\langle \overline{v},\overline{v}\rangle = 0$ if and only if $\overline{v}= \overline{0}$
Now let's observe equation
$$ Lf = u, $$
where $L$ is linear transformation in space $\mathcal{F}$ and $f,u \in \mathcal{F}$. Then weak form of this equation is
$$ \langle Lf, v \rangle = \langle u , v \rangle \quad \forall v \in \mathcal{F} $$
Proof that the equation $Lf=u$ guarantees satisfiability of $\langle Lf,v \rangle = \langle u , v \rangle$.
Could someone give an explanation/hint on how should I approach this? To put in other words, I am completely lost with this. I tried to take a scalar product with $v$ from both sides of the equation which leads to the equation with the scalar product. Then your somehow suppose to show that the equation is true $\forall v \in \mathcal{F}$ but not quite sure how to do this? After all, I'm not even sure if this is the right approach.