Prove weak formulation of linear system of equations

Question

Let $A: V\to V$ be a linear mapping and $Af=u$ where $u,f\in V$.

The weak formulation of $Af=u$ is $\langle Af,v \rangle$=$\langle u,v \rangle$ for all $v\in V.$

1. Prove that if $Af=u$ holds, then $\langle Af,v \rangle$=$\langle u,v \rangle$.

If I understood correctly, I have to prove that the inner product properties hold. Since $Au=f$ holds, then $Af-u=0$. Now, prove that

a) $\langle Af,v \rangle$=$\langle v,Af \rangle$

b) $\langle Af-u,v \rangle = \langle Af,v\rangle + \langle u,v\rangle$

c) $\langle c(Af-u),v\rangle=c\langle Af-u,v\rangle$

d) $\langle Af-u,Af-u\rangle \ge 0,$ and $\langle Af-u,Af-u\rangle =0 $ if and only if $Af-u=0.$

I know how to prove all of them except b) but on the other hand, I'm also quite unsure if I even understood the exercise right. The second part of the task asks

2. Let's assume $\langle Af,v \rangle$=$\langle u,v \rangle$ holds for some $v\in V$. Does this also mean that $Af=u$ holds?

I don't know to approach this at all. Should I somehow start with the inner product properties?

Answer 1

I think that you have misunderstood the first part of the problem. We are given a space $V$ that comes with an inner-product $\langle \cdot, \cdot \rangle$ over that space. The proof of 1, then, is very simple ("trivial", perhaps).

Suppose that $Af = u$. Then for all $v \in V$, we have $$ \langle Af,v \rangle = \langle u,v \rangle. $$

For the second part: the answer to the question is no. We can construct a counterexample with $V = \Bbb R^2$ as follows: take $A$ to be the identity map, take $f = (1,0)$, take $u = (2,0)$, and take $v = (0,1)$.