Let ||.||_1 and ||.||_2 be norms on a vector space X
such that X_1 = (X, ||.||_1) and X_2=(X,||.||_2) are complete.
If ||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0,
show that convergence in X_1 implies convergence in X_2.
and conversely show that there are positive numbers
a and b such that for all x element of X,
a||x||_1 <= ||x||_2 <= b ||x||_1.
Oh, c'mon! If converges to x in ||.||_1, then converges to 0 in ||.||_1. But your condition is "||x_n||_1 ---> 0 always implies ||x_n||_2 -----> 0" so converges to 0 in ||.||2 from which converges to x in ||.||_2.
First part of the problem is perfectly explained by HallsofIvy.
Second part consists of proving that || ||_1 and || ||_2 are equivalent norms. But the first part implies that the identity
I: (X,|| ||_1)----------(X,|| ||_2)
is continuous at zero, and hence continuous. Since both spaces are complete, Open Mapping Theorem implies that I is an isomorphism, and we are done.