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What is the difference between strong and weak convergence?

I am reading "Introductory functional analysis" by Kreyszig and I dont appreciate the differences between the two.

Definition of strong convergence:

A sequence $(x_n)$ in a normed space $X$ is said to be strongly convergent if there is an $x \in X$ such that $$\lim_{n \to \infty}||x_n-x||=0$$

Definition of weak convergence:

A sequence $(x_n)$ in a normed space $X$ is said to be weakly convergent if there is an $x \in X$ such that $$\lim_{n \to \infty}f(x_n)=f(x)$$

I do not appreciate the differences between the two, does anyone have an example to highlight the differences?

How does the proof differ in showing if a sequences converges weakly or strongly?

user232183
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    consider the normed space of complex sequences with the $l^2$ norm ($|x| = \sqrt{\sum_k |x(k)|^2}$), and show that the sequence of sequences $(\delta_n)$ converges weakly to the $0$ sequence. – reuns May 03 '16 at 16:23
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    Paying attention to what $f$ is in the definition of weak convergence should help. It's been a while, but isn't $f$ supposed to be a linear functional? You should also determine whether the definition requires an $f$ or an arbitrary $f.$ These distinctions tend to make all the difference. – Chris Leary May 03 '16 at 16:23

3 Answers3

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It all boils down to realizing that it is natural (and interesting!) to consider different topologies on the same set $X$, each of which comes with a notion of convergence.

In this scenario we have on one hand the strong topology, i.e. the topology induced by norm on $X$, and the so called weak topology on the other, i.e. the coarsest topology for which every element in $X^*$ is continuous.

The first thing to realize is that $\tau_{\text{weak}} \subset \tau_{\text{strong}}$. Indeed,

strong implies weak: if $x_n \to x$ and $f \in X^*$, then by definition of $X^*$ we have that $f(x_n) \to f(x)$. In other words, strong convergence implies weak convergence, weakly closed implies (strongly) closed etc.

The other implication does not hold as shown by the following example:

Weak does not imply strong: consider the Banach space $\ell^p$, $1<p<\infty$ and let $e_i$ be the sequence $(0,\dots,1,0\dots)$, with $1$ in the $i$-th position. Then it is easy to show that $\{e_i\}$ converges weakly to $0$ in $\ell^p$ but not strongly.

Here you can find a proof that for finite dimensional spaces the two topologies coincide.

It is worth mentioning it is not true that the topologies are different if and only if $X$ is infinite dimensional. For a counterexample you can take a look at this question ($\ell^1$ is a weird space).

Community
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Giovanni
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A weak convergence is defined in an inner product while a strong convergence is defined in a norm.

A strong convergence is also a weak convergence but not vice versa.

user555794
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In an arbitrary Hilbert space $H$ if $e_n$ is an orthonormal sequence, then by Riesz Representation one can show that for every $f$ in the dual of $H$, $f(e_n)$ converges to $0$. But since this sequence is not Cauchy and $H$ is complete, it is not convergent.

M. A.
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Erenay Güner
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