Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am looking for an example of a strongly continuous function and an example of a weakly continuous function at a real number "a"

share|cite|improve this question
What are strongly and weakly continuous functions? – t.b. Sep 16 '12 at 3:39

I'll assume that you are using the following definitions:

$\tt{(Def)}$ Let $f: \mathbb R \to \mathbb R$ be a linear function. Then we call $f$ strongly continuous if $f$ is continuous with respect to the usual (Euclidean) topology.

$\tt{(Def)}$ Let $\mathbb R^\ast$ denote the set of all continuous linear maps from $\ell: \mathbb R \to \mathbb R$. Then we call $f: \mathbb R \to \mathbb R$ weakly continuous if it is continuous with respect to the weakest topology on $\mathbb R$ that makes all $\ell \in \mathbb R^\ast$ continuous.

Then strongly continuous is the same as continuous (in the usual sense) hence an example of a (strongly) continuous function would for example be $f(x) = x^2$.

An example of a weakly continuous function would be $f(x) = x$. This is linear and by definition, since the topology on the domain is the weakest topology such that all linear maps $\mathbb R \to \mathbb R$ are continuous (while the topology on the range is the usual (Euclidean) topology).

Since the weak topology is weaker (or weaker equals) than the Euclidean topology, every weakly continuous function is also strongly continuous. The converse is not true in general but is true if your spaces are finite dimensional: see here for a proof.

As you see: the notions of strong and weak continuity coincide for maps $\mathbb R \to \mathbb R$.

share|cite|improve this answer
Thanks for the answer. I am still confused. Some people say that strongly continuous functions include $f(x) = a$, based on the following definition: $f$ is a strongly continuous function if and only if there exists a $\delta > 0$ for all $\epsilon > 0$ such that for all real $z$, if $0 < | z - a | < \delta$ then $| f(z) - f(a) | < \epsilon$. – user41624 Sep 18 '12 at 4:31
So, they classify $f(x) = x^2$ as a continuous function based on the following definitions: $f$ is continuous iff $\lim_{x\rightarrow a} f(x) = f(a)$, and $\lim_{x\rightarrow a} f(x) = f(a)$ iff for all $\epsilon > 0$ there exists a $\delta > 0$ such that all real $z$, if $0 < | z - a | < \delta\text{ then }| f(z) - f(a) | < \epsilon$. The definition of a weakly continuous function is $f$ is a weakly continuous function if and only if there exist an $\epsilon > 0$ and a $\delta > 0$ and for all real $z$, if $0 < | z - a | < \delta\text{ then }| f(z) - f(a) | < \epsilon$. – user41624 Sep 18 '12 at 4:31
@HansPedersen: I rather heavily edited your post. If I changed your meaning in any way, please feel free to revert my edits. – Rick Decker Sep 20 '12 at 1:03
@HansPedersen: So, weakly continuous only requires "exists $\epsilon>0$"? Then this property has nothing to do with continuity: it holds, for example, for $$f(x)=\begin{cases}1 \text{ if }x\in\mathbb Q; \\ 0 \text{ if }x\notin\mathbb Q.\end{cases}$$ Such "weak continuity" simply requires $f$ to be bounded on each finite interval. – user31373 Sep 20 '12 at 1:57
@HansPedersen: You should register your account such that the system can recognize you when you return. Then you'll be able to mark the answer as accepted instead of posting this answer-that-is-not-an-answer. – Henning Makholm Sep 20 '12 at 20:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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