How to prove that not all norms are equivalent in an infinite-dimensional vector space?

In particular, I would like to prove that for a space $X$ of continuous real-valued functions defined on interval $[0,1]$, every two norms $\|\ .\|_p$ ($p \in [1, \infty]$) are not equivalent.

  • 2
    $\begingroup$ ROUGH HINT - Use scaling. Take a localized function $f\in L^p$ and concentrate it at a point: $\lambda^{\frac1p}f(\lambda x)$. Let $\lambda \to \infty$. The $L^p$ norm is unchanged by this operation but the $L^q$ norm (with $q>p$) will explode. $\endgroup$ Feb 8 '16 at 12:09
  • $\begingroup$ Thank you for you comment. Could you please go into some more detail? $\endgroup$
    – pizet
    Feb 19 '16 at 15:31

Consider the two spaces $L^{p_1}(-1, 1), L^{p_2}(-1, 1)$ with $1\le p_1<p_2 < \infty$. Let $f \in L^{p_1}\cap L^{p_2}(-1, 1)$. For all $\lambda\ge 1$ define $$ f_\lambda(x)=\lambda^\frac1{p_1}f(\lambda x).$$ Then you have $$\tag{1} \|f_\lambda\|_{p_1}=\|f\|_{p_1} $$ and $$\tag{2} \|f_\lambda\|_{p_2} = \lambda^{\frac{p_2-p_1}{p_1p_2}}\|f\|_{p_2}$$ If the two norms were equivalent on $L^{p_1}\cap L^{p_2}$ you would have, by definition, $$ c\|f_\lambda\|_{p_2} \le \|f_\lambda\|_{p_1}\le C\|f_\lambda\|_{p_2}$$ but (1) and (2) show that this is not possible (to see why, let $\lambda \to \infty$).


Simple example of two non-equivalent norms on infinitely-dimensional space:

Consider space of all contnuously differentiable functions $X = C^1 [0,1]$. Then equipping it with the norm: $$ \|f \|_{C^1} = \sup \limits_{x \in [0,1]} |f| + \sup _{x \in [0,1]} |f'| $$ gives us a complete space (Banach space), but if we consider norm: $$ \|f \|_\infty = \sup \limits_{x \in [0,1]} |f| $$ Then the normed space is not complete, hence the norms are not equivalent.

Another way to see it is to consider sequence $f_n(x) = \frac 1n \sin (2 \pi n x)$, which is convergent to $0$ function in $\| \cdot \|_\infty$: $$ \| f_n \|_\infty = \frac 1n \rightarrow 0 $$ norm, but not convergent to $0$ in $ \| \cdot \|_{C^1}$: $$ \| f_n \|_{C^1} = \frac 1n + 1$$

  • $\begingroup$ But I would like to prove this for every two norms $\|\cdot\|_{p_1}$ and $\|\cdot\|_{p_2}$ such that $p_1, p_2 \in [1,\infty]$. You proved this for $p_1 = 0$ and $p_2 = 1$. $\endgroup$
    – pizet
    Feb 8 '16 at 12:52
  • $\begingroup$ No, its not what I did. I gave You totally different example, answer to question about two non-equivalent norms. Notice, that space $X$ is a space of all continuously differentiable functions. $\endgroup$
    – wroobell
    Feb 8 '16 at 12:55
  • $\begingroup$ For the second, more difficult task You have already got a hint. I wanted to show easier example, and provide ways for showing non-equivalence of norms. $\endgroup$
    – wroobell
    Feb 8 '16 at 13:00

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