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Show the continuous embedding $$ \ell^2 \subseteq c_0. $$


I couldn't find a question that showed this "simple" proof, and I'm having trouble doing it myself.


For normed spaces $\mathcal{X},\mathcal{Y}$, let $x \in \mathcal{X}$. Then \begin{equation} \mathcal{X} \subseteq \mathcal{Y} \iff \exists c > 0 \;:\;\Vert x \Vert_{\mathcal{Y}} \le c \Vert x \Vert_{\mathcal{X}} \quad \forall x\in\mathcal{X}. \end{equation}

We say normed space $\mathcal{X}$ continuously embeds into normed space $\mathcal{Y}$.

The sequence spaces $\ell^2,c_0$ have norms $$\Vert x \Vert_2 = \left( \sum_{n} |x_n|^2 \right)^{1/2}$$ and $$\Vert x \Vert_{c_0} = \sup_{n} |x_n|,$$ respectively.

We say $x \in \ell^2 \iff \Vert x \Vert_2 < \infty$ and $x \in c_0 \iff \lim_{n\to\infty} x_n = 0$.


My attempt:

Suppose $x \in \ell^2$. Then $\lim_{n \to \infty} |x_n| = 0$. Hence, there exists $k$ such that $x_k = \sup_{n} |x_n|$. Therefore,

\begin{align} \mathrm{LHS} & = \sup_{n} |x_n| \\ & = |x_k| \\ & \le \sum_{n} |x_n| \text{ since $|x_n| \ge 0$ for $n\ne k$} \\ & \le c \left( \sum_{n} |x_n|^2 \right)^{1/2} \end{align} since it is true that for all $u,v\in\mathbb{R}$, $u \ge 0$, $v \ge 0$, there exists $c \ge 0$ such that $u \le cv$. Hence, $\ell^2 \subseteq c_0$.


Is the property of real numbers that I used enough to justify the ending of my "proof"?

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    $\begingroup$ For $x\in l^2$, the sum $\sum\limits_{n}{|x_n|}$ might not even exist (e.g. $x_n = 1/n$). Instead, note that $|x_k| = (|x_k|^2)^{1/2}\le\left(\sum\limits_{n}{|x_n|^2}\right)^{1/2}$. $\endgroup$
    – Joey Zou
    Sep 28, 2016 at 4:18
  • $\begingroup$ @BrianM.Scott I don't want to show that if $x \in \ell^2$ then $x \in c_0$ (sorry, the question was badly worded). I wanted to show that there is a continuous embedding $\ell^2 \subseteq c_0$. To do this, I believe that I have to use the inequality from the definition I gave. $\endgroup$
    – jamesh625
    Sep 28, 2016 at 4:33

2 Answers 2

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If $x \in l_2$ then $x \in c_0$.

We have $|x_k| \le \|x\|_2$ for all $ k$, hence $\|x\|_{c_0} = \sup_k |x_k| \le \|x\|_2$.

Just take $c=1$.

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Question was since edited, rendering this answer obsolete. Here it was:


This is incorrect:

For normed spaces $\mathcal{X},\mathcal{Y}$, let $x \in \mathcal{X}$. Then \begin{equation} \mathcal{X} \subseteq \mathcal{Y} \iff \exists c > 0 \;:\;\Vert x \Vert_{\mathcal{Y}} \le c \Vert x \Vert_{\mathcal{X}} \quad \forall x\in\mathcal{X}. \end{equation}

I'm going to assume that you're talking about $\mathcal X$ and $\mathcal Y$ as vector spaces over the same field $\mathbb R$, in which case this is correct.


By definition of $\ell^2$, everything within has a norm:

$$||(x_1,x_2 , \dots )||_{\ell^2}= \sqrt{\sum_{i} x_i^2}$$ The sum converges, which requires that the individual terms of the sum converge to $0$: $$\lim_{i\rightarrow \infty} x_i^2 = 0.$$

This in turn requires that $$\lim_{i \rightarrow \infty} x_i = 0.$$

Note that this condition on $\mathbf x$ is both necessary and sufficient for $\mathbf x \in c_0$, as this is precisely how $c_0$ is defined.

Hence $$\forall \mathbf x \in \ell^2, \quad \mathbf x \in c_0.$$ $$\Rightarrow \quad \ell^2 \subseteq c_0$$

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    $\begingroup$ Won't you also explain the inequality $\| x\|_{c_0} \le C \|x\|_{\ell^2}$? $\endgroup$
    – user99914
    Sep 28, 2016 at 4:29
  • $\begingroup$ @JohnMa - I don't know what to say, other than that seems to me a non-sequitur. $\endgroup$
    – Myridium
    Sep 28, 2016 at 4:30
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    $\begingroup$ The symbol "$\subseteq$" is often used to denote not just inclusion, but also continuous inclusion or continuous embedding. This is the point of the question. $\endgroup$
    – Joey Zou
    Sep 28, 2016 at 4:31
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    $\begingroup$ @JoeyZou - In the same way that the symbol $\subset$ is often used to mean $\subseteq$? It's called abuse of notation. $\endgroup$
    – Myridium
    Sep 28, 2016 at 4:33
  • $\begingroup$ Yes, sorry I worded the question poorly... $\endgroup$
    – jamesh625
    Sep 28, 2016 at 4:34

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