# Why does $e_i \in \ell^2$ weakly converge to $0$?

In our lecture notes of functional analysis we are given the following definition:

Let $(X, \| \cdot \|_X)$ be a $\mathbb{R}$-vector space with a norm and $X^*$ be its dual space. Let $(x_k)_{k \in \mathbb{N}} \subset X$ and $x \in X$. The sequence $(x_k)_{k \in \mathbb{N}}$ converges weakly to $x$ if for all $\ell \in X^*$ we have $$\ell(x_k) \to \ell(x)$$ as $k \to \infty$.

We are given the example, that the sequence $\ell^2 \ni e_i = (0, \dots, 0, 1, 0, \dots)$ converges weakly to $0$. What exactly is $\ell^2$? Is it relevant? If it is not, why does this sequence converge weakly to $0$?

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$l^2$ is collection of square summable sequences... –  Praphulla Koushik Jan 7 '14 at 12:56
$\ell^2$ is the space of square-summable sequences, $$\ell^2 = \left\lbrace (x_k) : \sum_{k=0}^\infty \lvert x_k\rvert^2 < \infty \right\rbrace.$$ It is isomorphic to its own dual (as a Hilbert space), so every linear form on $\ell^2$ is given by $$\lambda_y \colon x \mapsto \sum_{k=0}^\infty x_k\cdot y_k$$ for some $y \in \ell^2$. Since $y_k \to 0$ for every $y\in \ell^2$, and $\lambda_y(e_n) = y_n$, we see that $e_n \xrightarrow{w} 0$. –  Daniel Fischer Jan 7 '14 at 12:57

What exactly is $\ell^2$?

See Wikipedia or Daniel Fischer's comment (copied below, since it's very much like an answer).

Is it relevant?

I initially understood this part as: "why do we care about $\ell^2$"?" To that I would answer: $\ell^2$ is a canonical form of separable Hilbert space; every infinite-dimensional separable Hilbert space is isometrically isomorphic to $\ell^2$. From the same Wikipedia page: "[Hilbert spaces] are indispensable tools in the theories of partial differential equations, quantum mechanics, Fourier analysis (which includes applications to signal processing and heat transfer)—and ergodic theory, which forms the mathematical underpinning of thermodynamics."

But perhaps you asked whether mentioning a space is relevant to the issue of a particular sequence being weakly convergent to $0$. The answer is: yes, because the definition of weak convergence makes references to the space $X$ and its dual $X^*$. In order to decide whether something converges weakly, you have to know what is the space being considered. For example, if the same sequence $e_i = (0, \dots, 0, 1, 0, \dots)$ is considered as a sequence in $\ell^1$ (which is a different space), then it does not converge weakly to anything.

why does this sequence converge weakly to $0$?

Answered in the comment by Daniel Fisher:

$\ell^2$ is the space of square-summable sequences, $$\ell^2 = \left\lbrace (x_k) : \sum_{k=0}^\infty \lvert x_k\rvert^2 < \infty \right\rbrace.$$ It is isomorphic to its own dual (as a Hilbert space), so every linear form on $\ell^2$ is given by $$\lambda_y \colon x \mapsto \sum_{k=0}^\infty x_k\cdot y_k$$ for some $y \in \ell^2$. Since $y_k \to 0$ for every $y\in \ell^2$, and $\lambda_y(e_n) = y_n$, we see that $e_n \xrightarrow{w} 0$.

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