# Product of two continous functions is continous

1. Jun 24, 2011

### l888l888l888

1. The problem statement, all variables and given/known data
prove that the product of continous functions is continous

2. Relevant equations

3. The attempt at a solution
f is continous for all a. given epilon>0 there exists a delta>0 st |x-a|<delta implies |f(x)-f(a)|< episilon.g is continous for all a. given epilon>0 there exists a delta2>0 st |x-a|<delta2 implies |g(x)-g(a)|< episilon.|f(x)g(x)-f(a)g(a)|=|f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jun 24, 2011

### micromass

Staff Emeritus
Hi l888l888l888!

Continue on this path. Apply the triangle inequality to get a sum of two absolute values...

3. Jun 24, 2011

### l888l888l888

im sorry. I forgot to add that step. What i got was...
|f(x)g(x)-f(x)g(a)+f(x)g(a)-f(a)g(a)|<=
|f(x)||g(x)-g(a)|+|g(a)||f(x)-f(a)|. This was where I got stuck. But I do know from the assumptions that |g(x)-g(a)| and |f(x)-f(a)| are less that epsilon. so we get...
|f(x)||g(x)-g(a)|+|g(a)||f(x)-f(a)|<|f(x)|*epsilon+|g(a)|*epsilon?

4. Jun 24, 2011

### micromass

Staff Emeritus
Yes, the only problem lies with |f(x)| which is still dependent of x. But we know that

$$|f(x)-f(a)|<\varepsilon$$

Can we deduce something about $|f(x)|$ this way?

5. Jun 24, 2011

### l888l888l888

What do you mean. the only way I can think of seperating |f(x)| is...
epsilon>|f(x)-f(a)|>=|f(x)|-|f(a)|. but i dont even think that works because you have to put another absoulte vale around that last term????

6. Jun 24, 2011

### micromass

Staff Emeritus
Think of

$$|f(x)|=|f(x)-f(a)+f(a)|$$

7. Jun 24, 2011

### l888l888l888

Ok so |f(x)|=|f(x)-f(a)+f(a)|<=|f(x)-f(a)|+|f(a)|<epsilon + |f(a)|

8. Jun 24, 2011

### micromass

Staff Emeritus
Indeed, so that gives you what you're looking for.

9. Jun 24, 2011

### l888l888l888

10. Jun 24, 2011

### micromass

Staff Emeritus
Well, you've proven now that

$$|f(x)g(x)-f(a)g(a)|\leq \epsilon^2+\varepsilon |f(a)|+\varepsilon |g(a)|$$

Thus the left-hand side can be made arbitrarily small. And this is what you want...

11. Jun 24, 2011

### l888l888l888

OHHH. ok thanks!!!!

12. Jun 25, 2011

### l888l888l888

a minor detail that i overlooked was that f and g are continous real valued functions on a metric space M. needed to show fg is continous on M also. Does this change the outcome?

13. Jun 25, 2011

### micromass

Staff Emeritus
No. You can show in general that the multiplication map

$$\cdot:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}:(x,y)\rightarrow x\cdot y$$

is continuous (using techniques above). Then the product of two functions is continuous as composition of continuous functions:

$$fg=\cdot(f,g)$$

You may wish to work this out in detail.

14. Jun 25, 2011

I'm not an expert in topology, but if your metric is not defined as d(x,y) = |x-y| then you'll need modifications because your definition of being continuous assumes that the metric is the Euclidean metric but the outcome will be the same.

15. Jun 25, 2011

### l888l888l888

the metric is not defined specifically. where will the modifications need to be placed, just in the assumption of continuity?

16. Jun 25, 2011

### micromass

Staff Emeritus
You just need to modify the first bit:

That is, you'll need to change the appropriate absolute values with the metric. However, the absolute value of things like |f(x)-f(a)| remain because you're working in $\mathbb{R}$ there.

17. Jun 25, 2011

I think the definition of continuity says that a function f:X->Y is continuous at the point x if for any neighborhood of fx in Y there exists a neighborhood of x that its image is contained in the neighborhood of fx. in other words, fN(x,delta,X) must be a subset of N(fx,epsilon,Y). so, if your metric is defined to be the Euclidean metric, you'll have the old nice definition of continuity but if your metric is not Euclidean, then the calculations will be different because this time |x-a|<delta -> |f(x) - L|<epsilon is not necessarily what you have. I mean let our metric be something different and then this definition need to be modified to something that comes from our metric function.
Actually I'm a freshman in pure mathematics and have no experience in Topology, so maybe It was stupid to comment on something that I don't know much about it but I think that when the metric is not Euclidean the Bolzano definition of continuity should be modified. but still, that won't change the outcome I guess.

18. Jun 25, 2011

### micromass

Staff Emeritus
The only thing that needs to be modified for real-valued functions is the first absolute value. So it becomes

$$\forall \varepsilon>0:\exists \delta>0:\forall x:~d(x,a)<\delta~\Rightarrow~|f(x)-f(a)|<\varepsilon$$

19. Jun 25, 2011