# Stronger, Weaker, and Equivalent Norms on $C[0,1]$.

This is a problem from a past qualifying exam for applied analysis.

Consider the $1$, $2$ and $\infty$-norms on the usual Banach Space of continuous functions on $[0,1]$.

1. Under which of these norms is $C[0,1]$ complete?

2. Determine which norms are stronger than which others. Are any of them equivalent?

Clearly, $C[0,1]$ is complete under the $\infty$-norm--I don't have trouble proving this. I'm relatively certain that it is not complete under the $L^1$ and $L^2$ norms, but I'm not exactly sure how to show this.

For 2, I'm a bit confused about what it means to show one norm stronger than another. I have a clear definition for two norms being equivalent, but does $||\cdot||_1$ is stronger than $||\cdot||_2$ just mean there is some constant $c>0$ such that $||x||_2 \leq c ||x||_1$ for all $x$ in the appropriate space?

For (1), it's enough to show that the norms are not equivalent: if a normed space is complete under two norms, the identity map is an isomorphism of Banach spaces, and this implies that the norms are equivalent. Or you could show directly that $C[0,1]$ is not complete under the $1$ or $2$ norms by exhibiting a member of $L^1$ or $L^2$ that is not continuous and showing that it is the limit in $1$ or $2$ norm of a sequence of continuous functions. A convenient example is the indicator function of $[0,1/2]$.
Yes, you have the correct definition of "stronger" for norms. You might want to use the Cauchy-Schwarz inequality to compare $L^1$ and $L^2$ norms.
• Any advice to show that $L^1$ and $L^2$ are not equivalent on $C[0,1]$? I used Cauchy Schwarz to show that $L^1$ is stronger, but still need to show that they are not equivalent. I'm assuming I need to find a sequence in $C[0,1]$ that converges in $L^2$ but not $L^1$? – Chriz26 Jan 1 '17 at 20:29
• Hint: there are powers $x^p$ that are in $L^1$ but not $L^2$. – Robert Israel Jan 2 '17 at 2:29