[Solved]: What is the difference between “Decision” and “Verification” in complexity theory?

The task of deciding membership is: given any input $x$, decide wether $x in L$, i.e. compute the following function: $qquad displaystyle chi_L(x) = begin{cases}1 &x in L 0 & xnotin Lend{cases}$ On the other hand, the task of verifying membership is: given any input $x$ and a (proposed) proof (or witness) of membership, check quickly wether $xin L$ by that proof¹. For example, consider prime factorisation. Given $n in mathbb{N}$, compute all prime factors of $n$. On the other hand, given $(n, {i_1, dots, i_k})$, verify that $prod_{j=1}^k i_j = n$. Which is easier? Another example: Given a weighted graph $G = (V,E)$, decide wether there is a Hamilton circle (that visits all nodes) with weight at most $k$. On the other hand, given $(G, (v_1, dots, v_n))$, verify wether the path $v_1 to dots to v_n$ visits all nodes exactly once and has weight at most $k$. Which is harder?

1. So you will say “no” if $x in L$ but the proof is wrong. That is fine, though, as we consider nondeterministic machines in this context; it is only important that we can guess the correct proof and verify it (quickly).