# Thread: Norms and Banch spaces

1. ## Norms and Banch spaces

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

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

Is it a Banach spae.

For 1. I can show the ---> implication. If they're equivalent there exists $M,m > 0: 0 \leq m || x_n - x || \leq || x_n - x||' \leq M ||x_n - x ||$ so if the sequence $(x_n)$ converges in $||.||$ then it must converge in $||.||'$. Analogously for the other way around.
For 2. I know that $C[0,1]$ is complete with the Supremum norm. Since $f'$ is cts. we can rewrite this as:

$||f|| = f(0) + ||g||_{\infty}$ where $f'(t) = g (t)$. 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.

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

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

Is it a Banach space.
You need to show that every Cauchy sequence converges in this norm. So let $(f_n)$ be Cauchy. Then $\|f_m-f_n\| = |f_m(0) - f_n(0)| + \|f_m'-f_n'\|_\infty \to0$ as $m,n\to\infty$. But if the sum of two positive quantities tends to 0 then so do both of them individually. Therefore $(f_n(0))$ is Cauchy in $\mathbb{R}$, and $(f_n')$ is Cauchy in $C[0,1]$ (with the sup norm). Since both those spaces are complete, there exist a real number c and a continuous function g such that $f_n(0)\to c$ and $f_n'\to g$.

Now define $f(x) = c + \int_0^x\!\!\!g(t)\,dt$ and show that $f_n\to f$ in the given norm.

3. Sorry I follow Opalg's idea in some sort of way....but how do you show f_n tends to f?

(feel really stupid for asking this!)

4. Originally Posted by vinnie100
Sorry I follow Opalg's idea in some sort of way....but how do you show f_n tends to f?
Use the fact that $f_n(x) - f_n(0) = \int_0^x\!\!\!f_n'(t)\,dt$ (fundamental theorem of calculus). Then $f(x) - f_n(x) = c-f_n(0) + \int_0^x\!\!\!(g(t)-f_n'(t))\,dt.$

5. Wow that makes perfect sense now and now I see why it is a Banach Space