BPCS-001 Why $\sum_k f(q)_k \delta q_k=0$ implies $f_k(q)=0$

Context:
While formulating variational principle, one frequently  encounters a statement that
\[ \sum_k f_k(q) \delta q_k=0 \]impiles that  \(f_k(q)=0\), because the variations \(\delta q_k\) are arbitrary. Statements like this appear for example in Hamilton's action principle, principle of virtual work etc.The proof is almost never given. We give proofs.

Proof 1:
Without loss of genarality, we assume that \(f_k\) are real.
Since (\delta q_k\)  are infinitesimmaly samll, otherwise arbitrary, we choose \[\delta q_k =\epsilon f_k(q).\] This gives \[ \sum_k f_k(q)^2 =0.\]  A sum  of squares can never be zero, unless  each terms is zero. This completes the proof.

Proof 2:
Since \(\delta q_k\) areindependent and arbtrary,  we choose \[\delta q_j= \begin{cases} \epsilon  & \text{for} q=k \\ 0  & \text{for} q\ne k \end{cases}.\] The given equation then implies \(\epsilon f_k =0\) and this  holds for each \(k\).