Product of two continous functions

In summary, we discussed the proof in a book that asks us to prove that the product of two continuous functions is continuous. We also considered the question of whether a function that is undefined at some point can have a product that is defined at that point. We provided a framework for the proof, using the fact that lim(x-->a)f(x)=f(a) and showing that |f(x+h)g(x+h)-f(x)g(x)| can be made smaller than any given epsilon.
  • #1
im2fastfouru
6
0
There is a proof in my book that asks us to prove that the product of two continuous functions is continuous. If anyone could help please reply back, thanks!
 
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  • #2
If h(x) is undefined at some point c (and thus not continuous), will it ever have a product that is defined at that point?
 
  • #3
It is not clear to me what SticksandStones response has to do with the question!

im2fastfouru, you know, of course, that the product, fg, will be continuous at a as long as the lim, as x goes to a, of f(x)g(x) is equal to f(a)g(a). Therefore, you must prove that, given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex], then |f(x)g(x)- f(a)g(a)|< [itex]\epsilon[/itex]. Using, of course, the fact that f and g are both continuous at a, that is, that given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if |x-a|< [itex]\delta[/itex] such that |f(x)- f(a)|< [itex]\epsilon[/itex] and similarly for g.

I would recommend you look at |f(x)g(x)- f(x)g(a)+ f(x)g(a)- f(a)g(a)|.

After all, your book asked you to prove this, not us!
 
  • #4
I thought he was asking something entirely different; I apologize.:redface:
 
  • #5
hint
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
note
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
|f(x+h)-f(x)|,|g(x+h)-g(x)|<eps=max(eps1,eps2)
also recall
|a+b+c|<|a|+|b|+|c|
 
  • #6
lurflurf said:
hint
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
note
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
|f(x+h)-f(x)|,|g(x+h)-g(x)|<eps=max(eps1,eps2)
also recall
|a+b+c|<|a|+|b|+|c|

Why on Earth would he do so? Halls hints are quite straightforward, he actually almost did all of what was needed to do.>< All he needs to do is show that If(x)I does not get too large as x-->a(or sth), in other words the op just needs to show that f(x) is bounded, which is easy to show since lim(x-->a)f(x)=f(a), so any function who has a limit is bounded at that limit point. all he needs to do is let e-eplylon= 1, and it is easy to show that f(x) is bounded. If(x)I=If(x)-f(a)+f(a)I<=If(x)-f(a)I+If(a)I< 1 +If(a)I

For the OP: do you see how to go now? Like Halls said, you should give some effort on your own next time, if you want to receive more help. People here won't just give us answers(including me since i often ask for help).
 
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  • #7
lurflurf may has misread "continuous" as "differentiable".
 
  • #8
the point (of continuity) is simply that if two numbers are respectively near two other numbers, then the products are also near each other.

to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small.
 
  • #9
sutupidmath said:
Why on Earth would he do so? Halls hints are quite straightforward.
To solve the problem and because it is fun. I agree, I provided a different (though very slightly) view.
HallsofIvy said:
lurflurf may has misread "continuous" as "differentiable".
theorem f is continuous if and only if
lim_h->0 (f(x+h)-f(x))=0
where (for picking nits) we accept as implicit f(x) exist and limit_h->0 f(x+h) exist
mathwonk said:
the point (of continuity) is simply that if two numbers are respectively near two other numbers, then the products are also near each other.

to see this, let the numbers be a+h and b+k and compare the product of ab to that of (a+h)(b+k), when h and k are small.
A useful framework

we desire to show that

if
|f(x+h)-f(x)|<eps1
|g(x+h)-g(x)|<eps2
then
|f(x+h)g(x+h)-f(x)g(x)|<eps
write
f(x+h)=f(x)+[f(x+h)-f(x)]
g(x+h)=g(x)+[g(x+h)-g(x)]
thus
|f(x+h)g(x+h)-f(x)g(x)|=|[f(x+h)-f(x)]g(x)+f(x)[g(x+h)-g(x)]+[f(x+h)-f(x)][g(x+h)-g(x)]|
<|[f(x+h)-f(x)||g(x)|+|f(x)||g(x+h)-g(x)|+|f(x+h)-f(x)||g(x+h)-g(x)]|
<eps1|g(x)|+|f(x)|eps2+eps1*eps2
letting
eps1=(eps/3)/|g(x)| if |g(x)|>0
eps2=(eps/3)/max(eps1,|f(x)|) if |f(x)|>0
we easily see the result is true
 
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Related to Product of two continous functions

1. What is the definition of a product of two continuous functions?

The product of two continuous functions is a new function that is obtained by multiplying the outputs of two different continuous functions for every input value.

2. How do you determine if a product of two continuous functions is also continuous?

In order for the product of two continuous functions to be continuous, both individual functions must be continuous at the same point. This means that the limit of each function as the input approaches a certain value must exist and be equal.

3. Can the product of two continuous functions be discontinuous?

Yes, it is possible for the product of two continuous functions to be discontinuous. This can happen if one or both of the individual functions have a discontinuity at a certain point, or if the limit of the product function at that point does not exist.

4. What happens to the continuity of a product of two continuous functions if one of the functions is not continuous?

If one of the functions is not continuous, then the product function will also not be continuous. This is because a single point of discontinuity in one of the functions will affect the overall continuity of the product.

5. Are there any special properties of a product of two continuous functions?

Yes, there are a few properties that hold for products of continuous functions. For example, the product of two even or two odd functions will also be even, while the product of an even and an odd function will be odd. Additionally, the product of two periodic functions with different periods will not be periodic.

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