1. Show that two norms are equivalent iff the convergence of a sequence in the first norm is equivalent to its convergence in the second norm.

2. Let $\displaystyle C^1[0,1]$ be the space of all real valued functions on $\displaystyle [0,1]$ which have continuous derivatives and define the norm by:

$\displaystyle ||f|| = f(0) + sup|f'(t)| $.

Is it a Banach spae.

For 1. I can show the ---> implication. If they're equivalent there exists $\displaystyle M,m > 0: 0 \leq m || x_n - x || \leq || x_n - x||' \leq M ||x_n - x ||$ so if the sequence $\displaystyle (x_n)$ converges in $\displaystyle ||.||$ then it must converge in $\displaystyle ||.||'$. Analogously for the other way around.

For 2. I know that $\displaystyle C[0,1]$ is complete with the Supremum norm. Since $\displaystyle f'$ is cts. we can rewrite this as:

$\displaystyle ||f|| = f(0) + ||g||_{\infty}$ where $\displaystyle f'(t) = g (t)$. So the new norm differs from the sup norm by a constant and then somehow show that they are equivalent. Then using a lemma: two norms are equivalent iff the normed spaces associated with each are both Banach finish it off.