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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

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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
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.

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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

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