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I am struggling with the following question:

Is it possible to find uncountably many norms on $C[0,1]$ such that no two are Lipschitz equivalent?

I had thought about trying to define norms for each real number r>1:

$$\|f\|_r =\left( \int_0^1 f(x)^r \, dx \right)^{1/r}$$

but I think these are Lipschitz equivalent.

Does anyone have any better ideas?

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  • $\begingroup$ "but I don't think these are Lipschitz equivalent" <- But that is exactly what you want, isn't it? That no two of the norms are (Lipschitz) equivalent. [And what you wrote only gives you a norm for $r \geqslant 1$. But there are still uncountably many of those.] $\endgroup$ – Daniel Fischer Nov 6 '15 at 12:52
  • $\begingroup$ Sorry, I meant to say I thought they were Lipschitz equivalent. Maybe they aren't? I can't find counterexamples though. $\endgroup$ – David Smith Nov 6 '15 at 13:54
  • $\begingroup$ No two of these norms are (Lipschitz) equivalent. Hölder's inequality gives $\lVert f\rVert_r \leqslant \lVert f\rVert_s$ for $r < s$, so you need to show that you can't estimate $\lVert f\rVert_s \leqslant C\cdot \lVert f\rVert_r$ then. Play with a modification of $f_s(x) = x^{-1/s}$ (you need to modify that to get a continuous function). $\endgroup$ – Daniel Fischer Nov 6 '15 at 14:24
  • $\begingroup$ Sophisticated answers! A lazier answer that might work (after I get my morning coffee) is $$\|f\|_r= \sup_{0\leq x\leq r}|f(x) | + \int_r^1|f(x)|\,dx$$ for $0<r<1$. It doesn't much to be a norm, although it takes a lot to be a useful norm. $\endgroup$ – B. S. Thomson Nov 6 '15 at 17:25
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In addition to the norms suggested in comments, here is one more (inspired by B.S.Thomson): for $t\in [0,1]$, let $$ \|f\|_t = |f(t)| + \int_0^1 |f(x)|\,dx $$ The integral term is only needed to make this a norm rather than a seminorm. The fact that these are mutually nonequivalent follows by considering $f_{a,n}(x)=\max(0,1-n|x-a|)$ which satisfies $$ \|f_{a,n}\|_t \to 0,\quad t\ne a;\qquad \text{yet } \ \|f_{a,n}\|_a = 1 $$

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