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"?

  • 3
    $\begingroup$ 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 '16 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 '16 at 4:33

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 $c=1$.


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

  • 2
    $\begingroup$ Won't you also explain the inequality $\| x\|_{c_0} \le C \|x\|_{\ell^2}$? $\endgroup$ – user99914 Sep 28 '16 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 '16 at 4:30
  • 2
    $\begingroup$ 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 '16 at 4:31
  • 1
    $\begingroup$ @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 '16 at 4:33
  • $\begingroup$ Yes, sorry I worded the question poorly... $\endgroup$ – jamesh625 Sep 28 '16 at 4:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.