Product of two continous functions is continous

[l888l888l888]

Homework Statement

prove that the product of continuous functions is continous

Homework Equations

The Attempt at a Solution

f is continuous for all a. given epilon>0 there exists a delta>0 st |x-a|<delta implies |f(x)-f(a)|< episilon.g is continuous 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)|...

 

[micromass]

Hi l888l888l888!

Homework Statement

prove that the product of continuous functions is continous

Homework Equations

The Attempt at a Solution

f is continuous for all a. given epilon>0 there exists a delta>0 st |x-a|<delta implies |f(x)-f(a)|< episilon.g is continuous 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)|...

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

 

[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?

 

[micromass]

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?

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?

 

[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 don't even think that works because you have to put another absoulte vale around that last term?

 

[micromass]

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 don't even think that works because you have to put another absoulte vale around that last term?

Think of

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

 

[l888l888l888]

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

 

[micromass]

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

 

[l888l888l888]

Please elaborate...

 

[micromass]

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...

 

[l888l888l888]

OHHH. ok thanks!

 

[l888l888l888]

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

 

[micromass]

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

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.

 

[AdrianZ]

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.

 

[l888l888l888]

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.

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

 

[micromass]

You just need to modify the first bit:

f is continuous for all a. given epilon>0 there exists a delta>0 st d(x,a)<delta implies |f(x)-f(a)|< episilon.g is continuous for all a. given epilon>0 there exists a delta2>0 st d(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)|...

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.

 

[AdrianZ]

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.

 

[micromass]

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.

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

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

 

[AdrianZ]

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

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

Yea. That's exactly what I objected to. It should be modified depending on what metric function we're using. also |f(x)-f(a)| should become d(f(x),f(a)) I think.