Proving $(f^*)^*=f$ I solved it like this : $$\langle (f^*)^*(v),w \rangle=\langle v,f^*(w)\rangle=\langle f(v),w\rangle$$ My lecture notes gave a proof with some more steps. Now i'm not sure, maybe i messed something.
here $f^*$ denotes the adjoint  of the linear map $f$.
 A: Once you have $\langle (f^*)^*(v),w \rangle=\langle f(v),w\rangle$ for all $v$ and $w$ you still need to do a little bit more to show $(f^*)^* = f$. First rewrite it as $\langle ((f^*)^* - f)(v),w \rangle=0$ which is true for all $v$ and $w$. In particular it is true for $w = ((f^*)^* - f)(v)$. So we get $\langle ((f^*)^* - f)(v),((f^*)^* - f)(v) \rangle=0 \implies ((f^*)^* - f)(v) = 0$ for all $v$. And so $(f^*)^* - f = 0 \implies (f^*)^* = f$.
A: From your argument, you can conclude that, for $y = (f^{\ast})^{\ast}(v) - f(v)$, one has
$$
\langle y,w \rangle = 0 \quad \forall w
$$
And you want to conclude that $y = 0$. For this you need to know that these linear functionals
$$
\varphi_w : x \mapsto \langle x,w\rangle
$$
separate points in your vector space. For this you will need the Riesz representation theorem and the Hilbert space version of the Hahn-Banach theorem. Is that what your book uses?
A: I assume we are working in a finite dimensional inner product space V. What your proof is missing is the well known fact that if $f:V \to V$ and $g:V \to V$ are linear transformations with $<f(v),w>=<g(v),w>$ for all $v,w \in V$, then $f=g$.  To prove this fact, let $\{e_1, ..., e_n\}$ be an orthonormal basis of $V$ (you can construct one via the Gram-Schmidt process). Then for each $v \in V$ we have the identity $v= \sum_{i=1}^n<v,e_i>e_i$. In particular $$f(e_j) = \sum_{i=1}^n <f(e_j), e_i>e_i = \sum_{i=1}^n<g(e_j),e_i>e_i = g(e_j)$$ Thus, $f$ and $g$ agree on a basis, so they are equal.
