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The following is a homework question and I'd be glad if you could tell me if I did it right:

Question:

Consider the space of real polynomials $\mathbb{R}[X]$. Define the family of norms $$ || p ||_s := || p ||_{C([0,s])} = \sup_{x \in [0,s]} |p(x)|$$ for any $s \in \mathbb{R}_{>0}$. Use the Stone-Weierstrass theorem to show that these are all inequivalent.

Answer: They are not actually all equivalent. There are two cases:

(i) $s \in (0,1)$:

Then $\sup_{x \in [0,s]} |p(x)| \in \mathbb{R}$ so $\exists K : |p(x)| \leq K$.

So for $s, s^\prime$, there exist $K, K^\prime $ such that $$ || p ||_s K \leq || p ||_{s^\prime} \leq || p ||_s K^\prime$$

(ii) $s \geq 1$:

Consider $p(x) = x^n$. Then $$ \lim_{n \rightarrow \infty} || x^n ||_s = \lim_{n \rightarrow \infty} s^n = \infty$$ So there are no $K,K^\prime$ such that the above holds.

I didn't use Stone-Weierstrass, so the answer is probably wrong but I don't see where. Many thanks for your help!

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Indeed, it seems there is a problem: $\lVert x^n\rVert_s=s^n$ hence we can't find $K$ such that $\lVert x^n\rVert_{s'}\leq K\lVert x^n\rVert_s$ if $s'>s$ (and the norm are never equivalent). –  Davide Giraudo Oct 1 '11 at 13:31
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There is something terribly wrong with that exercise and I don't see what the intention could be, as Stone-Weierstrass doesn't apply to $\mathbb{R}[X]$. As you notice, the sequence $p_n(x) = (x/s)^n$ converges to $0$ in the norm $\|\cdot\|_{s'}$ for every $s' \lt s$ while for $s' \gt s$ we have $\|p_n\|_{s'} \to \infty$. –  t.b. Oct 1 '11 at 13:32
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I didn't see the title where you speak about INequivalent norms, while the question in the body is about equivalence of norms. Edit: there was a comment asking whether I implied that the answer is correct No, I agree with David's view in the answer below, that in (i) you messed up. But the idea in (ii) generalizes easily, as I said in my previous comment. As David, I have no idea how Stone-Weierstrass is supposed to enter the picture here. –  t.b. Oct 1 '11 at 13:39
    
I made a typo, it should've been *in*equivalent! Sorry : ( –  Matt N. Oct 1 '11 at 13:39
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I think the idea with Stone-Weierstrass is this. To show the norms are inequivalent, one should produce polynomials which are small everywhere on $[0,s]$ but large somewhere on $[0,s']$ for $s' > s$ (i.e. large somewhere in $[s,s']$). You could use Stone-Weierstrass (actually just Weierstrass) for this: take any continuous function which has this property, and then Weierstrass guarantees there is a polynomial which is uniformly close to it. However, this is overkill, since it is very easy to just write down a polynomial with the desired property. –  Nate Eldredge Oct 1 '11 at 14:21
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1 Answer

up vote 2 down vote accepted

In your argument for case (i), your K and K' will depend on p, which is not allowed -- in the definition of equivalence of norms, you need to have constants which work for all vectors in the space at once. And I'm not sure what you're doing in case (ii) -- you make a statement about a sequence tending to $\infty$ in a specific norm, but how does that tell you anything about whether that norm is equivalent to some other one?

But I think that the question is totally messed up. I think these norms are all inequivalent. It suffices to show that for any distinct non-negative reals $s < t$, there is a sequence $f_n$ of polynomials that tends to 0 in the s-norm but not the t-norm. Can you see how to do this? (Edit: The comment by t.b. tells you how to do this.)

(Goodness knows what Stone--Weierstrass has to do with this. Are you sure that was the exact statement of the question?)

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Thank you! Yes, except for the typo where I left out the in in front of the equivalent. –  Matt N. Oct 1 '11 at 13:41
    
You can look at the homework sheet here –  Matt N. Oct 1 '11 at 13:43
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