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.

I have a homework question, to prove that if $f(x)$ is continuous at $x_0 = 0$ and $f(0) = 0$ then $g(x)$ is continuous at $x_0=0$, where $g(x)=f(x)D(x)$ and $D(x)$ is the Dirichlet Function.

I am having trouble proving this, could anyone please help me out?

Thanks a-lot

share|improve this question
1  
I'm not sure "the Dirichlet function" has one standard definition - it would help to include the definition in your question. –  Greg Martin Nov 25 '11 at 21:40

1 Answer 1

up vote 1 down vote accepted

You need that $f(0)=0$ otherwise you could take $f(x)=1$ which is continuous at $x_0=0$ but you have that $f(x)\cdot D(x)=D(x)$ which is not continuous at $0$.

If you have it however it is easy to see that is correct by using the fact that $|D(x)|\leq1$ everywhere. Therefore you have that $f(x)D(x)$ has to be $0$ at $x_0=0$ (I let you write out the details).

Edit: As you are asking for more details, I will give you another hint but the question is very trivial so I fear to completely solve it...

For $g(x)$ to be continuous at $0$ it is sufficient to show that

$$\lim_{x \rightarrow 0}g(x)=g \Big(\lim_{x \rightarrow 0}x \Big)=g(0)=f(0)D(0)=0.$$

Therefore you can use that

$$\lim_{x \rightarrow 0}|g(x)|=\lim_{x \rightarrow 0}|f(x)D(x)|\leq \cdots \leq 0.$$

Then your theorem follows, now you have to fill out the dots.

share|improve this answer
    
Oh sorry I forgot to write that $f(0)=0$. I still don't understand why $f(x)D(x) $ is continuous though.. –  Jason Nov 25 '11 at 19:07
    
@Jason I added some more notes, let me know if you get it now. –  Listing Nov 25 '11 at 19:15
    
Don't you need the limit of $D(x)$ to exist to do that? (sorry I might just be tired and not thinking straight) –  Jason Nov 25 '11 at 19:21
    
No only the limit of the product needs to exists, and it does exist. –  Listing Nov 25 '11 at 19:23
    
Don't $f(x)$ and $D(x)$ have to have limits to say that the limit of there product is the product of their limits? Or are you using something else here which I am blindly missing... I feel like I am missing something easy. –  Jason Nov 25 '11 at 19:37

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.