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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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    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.2011-10-01

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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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    You can look at the homework sheet [here](https://www.math.ethz.ch/education/bachelor/lectures/hs2011/math/fa/FAI_HS11_Serie2.pdf)2011-10-01