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Let $p\in [1,2)$ and define $B=\{(x_n)_n: \sum_n |x_n|^p\leq R\}$ for any $R>0$. I need to show that this set is closed in $\ell^2$.

My idea was the following.

Let me pick $(x^{(k)})_k=((x_n^{(k)})_n)_k\in B$ a convergent sequence, say $((x_n^{(k)})_n)_k\rightarrow (x_n)_n$ as $k\rightarrow \infty$. I need to show that $(x_n)_n\in B$. Hence consider $$\begin{align} \sum_n |x_n|^p&=\sum_n \left|\left(x_n-x_n^{(k)}\right)+x_n^{(k)}\right|^p\\&\leq\sum_n \left(\left|\left(x_n-x_n^{(k)}\right)\right|+\left|x_n^{(k)}\right|\right)^p \end{align}$$

Now the problem is that here I need to apply the Binomial Theorem to continue, hence I wanted to ask if there is an easier way to find an upper bound. I want to separate this two summands since I know that $\sum_n |x_n^{(k)}|^p\leq R$ and $\sum_n \left|\left(x_n-x_n^{(k)}\right)\right|^p $ can be bounded since $((x_n^{(k)})_n)_k\rightarrow (x_n)_n$.

Could maybe someone give me a hint how to proceed?

user123234
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  • I think you can use convexity argument: $f(t)=t^p$ is convex for $t>0$, so $(\frac{a+b}{2})^p\le \frac{a^p+b^p}{2}$, $a,b\ge 0$. – Vit Nov 13 '22 at 22:08
  • @Vit I'm not convinced yet, because as I remember, your property is called midpoint convex and where should I get my $\frac{\cdot}{2}$? – user123234 Nov 13 '22 at 22:22
  • Well, $|x_n|^p=2^p|\frac{x_n}{2}|^p$. – Vit Nov 13 '22 at 22:25
  • @Vit But let me pick $a=0.2$ and $b=1.5$ and $p=0.5$, then $\left(\frac{a+b}{2}\right)^p=0.92195$ but $\frac{a^p+b^p}{2}=0.57716$ so $t^p$ is not midpoint convex or am I wrong? – user123234 Nov 13 '22 at 22:38
  • I assumed $p>1$. – Vit Nov 13 '22 at 22:39
  • I see thanks a lot – user123234 Nov 13 '22 at 22:44
  • @Vit but at the end I get $2^{p-1}||x_n-x_n^{(k)}||_2^2+R2^{p-1}$. I know that the first summand converges to $0$ as $k\rightarrow \infty$ but then I still have $R2^{p-1}$ which is not less or equal then $R$ – user123234 Nov 13 '22 at 22:54
  • I misunderstood your "for any $R>0$". I thought you meant for all finite values $R$, not some fixed one. Anyway, I don't see how you can get the bound with this method then since you assume that it's possible to have $\sum_n x_n^{(k)}= R$ and $|(x_n-x_n^{(k)})|>0$. – Vit Nov 13 '22 at 23:12
  • $B$ is closed because its the continuous preimage of a closed set – C Squared Nov 14 '22 at 00:24
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    The function $f(x)=|x|p=\left (\sum{n=1}^\infty |x_,n|^p\right)^{1/p}$ satisfies $$|f(x)-f(y)|\le f(x-y)\le |x-y|_2.$$ Hence it is continuous. The set is if the form ${x\in\ell^2:, f(x)\le R}.$ Thus the set is closed. – Ryszard Szwarc Nov 14 '22 at 01:02
  • Echoing @RyszardSzwarc, for $1 \le p \le q \le \infty$ we have $|x|_p \le |x|_q$ (on the $l_r$ spaces, that is). Hence if $x_n \to x$ in $l_2$ then $x_n \to x$ in $l_p$ since $p \in [1,2)$. – copper.hat Nov 14 '22 at 03:45
  • @RyszardSzwarc for $p\le 2$, isn't the inequality in the other direction ? See this post for instance, we have $$|x|_q\le|x|_p $$ when $p\le q$ – Stratos supports the strike Nov 14 '22 at 08:19
  • @StratosFair Right!! Thanks. My answer is correct for $p>2.$ The funny thing is that I have mislead copper hat – Ryszard Szwarc Nov 14 '22 at 08:25
  • @copper.hat My comment is wrong. Sorry for the misleading. – Ryszard Szwarc Nov 14 '22 at 08:34
  • This follows directly from Fatou's Lemma. – daw Nov 14 '22 at 08:50
  • Another reminder not to drink and derive! – copper.hat Nov 14 '22 at 14:21
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    @copper.hat For me: not to watch ATP finals while answering questions. – Ryszard Szwarc Nov 14 '22 at 16:00

1 Answers1

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The answer in my comment is wrong (thanks @StratosFair). Let $R>0.$ Assume $x^{(n)}\in B$ and $\|x^{(n)}-x\|_2\to 0.$ Thus $x^{(n)}_k\to x_k$ for any $k.$ Then for any fixed $K$ we have $$\sum_{k=1}^K|x_k|^p=\lim_{n\to \infty}\sum_{k=1}^K|x_k^{(n)}|^p\le R$$ Hence $$\sum_{k=1}^\infty |x_k|^p\le R$$ Thus $x\in B.$

Ryszard Szwarc
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