all norms are not equivalent

Let $V$ be the vector space of continuous functions $[-1,1]\to ℝ$ that are differentiable at $0$. Then we can define norms

$$\parallel f\parallel =\underset{x\in [-1,1]}{\max }|f|,$$

and

$${\parallel f\parallel }^{\prime }=\parallel f\parallel +|f^{\prime }(0)|.$$

It is not difficult to find a sequence of functions $f_1,f_2,\mathrm{\dots }$ in $V$ such that

1. $f_k^{\prime }(0)=k$ for $k=1,2,\mathrm{\dots }$,

2. $\parallel f_k\parallel =1$.

Then $\parallel f_k\parallel =1$, and ${\parallel f_k\parallel }^{\prime }=1+k$, so there is no $C>1$ such that

$${\parallel f\parallel }^{\prime }\le C\parallel f\parallel \mathit{\hspace{1em}}f\in V,$$

and $\parallel \cdot \parallel$ and $\parallel \cdot {\parallel }^{\prime }$ cannot be .

Title	all norms are not equivalent
Canonical name	AllNormsAreNotEquivalent
Date of creation	2013-03-22 15:36:11
Last modified on	2013-03-22 15:36:11
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	6
Author	matte (1858)
Entry type	Example
Classification	msc 46B99