Let $Q$ be an operator and $|f\rangle$ a vector in a complex Hilbert space H.
How do I prove that
1. If $\langle f|Q|f\rangle=0$ for every $|f\rangle$ in H, then $Q$ is zero;
2. $Q$ is hermitian iff $\langle f|Q|f\rangle\in \mathbb{R}$ for every $|f\rangle$ in H?
1 Let's write $|f\rangle$ as $\sum_i a_i|e_i\rangle$. Then $\langle f|Q|f\rangle=\sum_j a_j\langle e_i|Q\sum_i a_i|e_i\rangle=\sum_j\sum_i a_ja_i\langle e_j|Q|e_i\rangle=0$, so $\langle e_j|Q|e_i\rangle=0$ for all $i,j$. Does this immediateluy imply that $Q=0$?
2 If $Q$ is hermitian, then $Q^\dagger=Q$. This means that $\langle f|Q|f\rangle=\langle Q^\dagger f|f\rangle$ How do I continue from here? If $|f\rangle$ were an eigenvector it would be easy, but the statement must hold for all $|f\rangle$.
If $\langle f|Q|f\rangle\in \mathbb{R}$, I am supposed to use 1. but I don't see how.
Hint for 1: One approach is to use the polarization identity. Another approach is to consider the expressions $$ \langle f+g | Q | f+g \rangle , \qquad \langle f+ig | Q | f+ig \rangle $$
Whichever way you approach the problem, it is helpful to note that if $\langle f | Q |g\rangle$ for every combination of vectors $|f\rangle,|g\rangle$, then $Q$ is necessarily $0$.
Hint for 2: The actual statement (which you seem to have copied incorrectly) is as follows:
If $Q$ is Hermitian, it suffices to note that for any $Q$, we have $$ [\langle f | Q|f \rangle]^* = [\langle f | Q|f \rangle]^\dagger = \langle f|Q^\dagger |f \rangle $$ where $^*$ denotes the complex conjugate. For the other direction: if $\langle f |Q |f\rangle \in \Bbb R$ for all $|f \rangle$, what can you say about $\langle f | (Q - Q^\dagger)|f \rangle$?