Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X = \mathbb{R}$ and $X^\mathbb{R}$ denote the set of functions $\mathbb{R} \to \mathbb{R}$. Let $B \subseteq X^\mathbb{R}$ denote the subset of bounded functions $\mathbb{R}\to \mathbb{R}$, i.e. the set of those functions $f$ for which there exists some $M_f$ such that $f(x) < M_f$ for all $x$. Show that $(f,g) \mapsto fg$ does not give a continuous function $X^\mathbb{R} \times X^\mathbb{R}\to X^\mathbb{R}$ when $X^\mathbb{R}$ is equipped with the uniform topology but that the restriction $B \times B \to B$ is continuous.

I'm trying the construct a counterexample using the sequence limit definition of continuity. I understand that for the first part, I am looking for an unbounded function where the preimage of $U$ open in $X^\mathbb{R}$ with the uniform topology, meaning it's not within a distance of $1$, is not open. But I can't think of an explicit example.

share|improve this question
2  
As a general tip, if $X=\mathbb R$, just use $\mathbb R$. No need for additional notations; the word "belongs" is ambiguous: from the context it is clear you mean subset, but it could mean "a member of". –  Asaf Karagila Sep 30 '11 at 4:40

1 Answer 1

Let $f$ be any unbounded function that is never $0$. Let $g$ be its inverse with respect to multiplication, i.e., $g(x)=1/f(x)$. Now $fg$ is constantly $1$, but multiplication is not continuous at $(f,g)$. To prove this, use the $\epsilon$-$\delta$-definition of continuity rather than the sequential definition.


From Asaf's comments I conclude that I was not very clear. I will fill in some details. I assume that by "uniform topology" you mean the topology generated by sets of the form $$U_{\varepsilon}(h)=\{e\in\mathbb R^{\mathbb R}:\sup_{x\in\mathbb R}|e(x)-h(x)|<\varepsilon\}$$ where $h:\mathbb R\to\mathbb R$ and $\varepsilon>0.$

Now let $f$ and $g$ be as above, i.e., $f$ unbounded and never $0$, $g$ its pointwise inverse. Let $\varepsilon=1$. $fg$ is constantly $1$. I will show that there is no $\delta>0$ such that for all $f'\in U_\delta(f)$ and all $g'\in U_\delta(g)$ we have $f'g'\in U_\varepsilon(fg)$.

Namely, let $\delta>0$. Let $x\in\mathbb R$ be such that $|f(x)|>10/\delta$. Now $|g(x)|<\delta/10$. Choose $h\in U_\delta(g)$ such that $h(x)>\delta/2$. We have $|f(x)h(x)|>10/2=5$. In particular, $fh\not\in U_\varepsilon(fg)$. This shows that pointwise multiplication is not continuous on $\mathbb R^{\mathbb R}$ with the uniform topology.

share|improve this answer
    
Stefan, are you sure you are reading this correctly? I'd think that the function of pointwise multiplication $(f,g)\mapsto fg$ is not continuous; furthermore $\mathbb R^\mathbb R$ is not metrizable so $\epsilon-\delta$ is meaningless there. –  Asaf Karagila Sep 30 '11 at 8:47
    
Asaf, the topology on $\mathbb R^{\mathbb R}$ that we are talking about is the uniform topology, so I assume that means generated by $\varepsilon$-balls in the sup-"metric". –  Stefan Geschke Sep 30 '11 at 13:03
    
Moreover, I am claiming that pointwise multiplication is not continuous, just like you say. –  Stefan Geschke Sep 30 '11 at 13:22
    
Hi, thanks, that's really helpful. –  jamie coulter Sep 30 '11 at 13:38
    
Stefan, actually the answer was quite clear. It was the question I found unclear, making me uncertain that you answered the actual question. However, if the OP says that you've been helpful - who am I to judge? –  Asaf Karagila Sep 30 '11 at 15:08

Your Answer

 
discard

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.