so I have $T:V->V$ with $V =L2([0,1])$.
$$f->(Tf)$$
And $$Tf(x)=xf(x)$$ for all $x\in [0,1]$.
And I have to prove $T$ is well defined, the thing is I have no idea how to progress, I've been reading similar questions that have been answered here, and I get that I'm supposed to prove that for some $a\in V$ there should be some $b$ and $b'$ such that $(a,b)$ $=$ $(a,b')$.
And that I cant use $f(x)$ because I haven't proved yet that it's a function, but then how would I get to the equality?
The only thing you need to do is to prove that if $f\in L^2\bigl([0,1\bigr)$, then $x\mapsto xf(x)\in L^2\bigl([0,1\bigr)$.