# Product of two continous functions is continous

• l888l888l888
In summary, the conversation discusses the proof that the product of continuous functions is continuous. The participants use the triangle inequality and the definition of continuity to show that the left-hand side of the equation can be made arbitrarily small, regardless of the metric used. They also mention that if the metric is not defined as the Euclidean metric, some modifications may need to be made to the definition of continuity. However, this does not affect the outcome of the proof.
l888l888l888

## Homework Statement

prove that the product of continuous functions is continous

## 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)|...

Hi l888l888l888!

l888l888l888 said:

## Homework Statement

prove that the product of continuous functions is continous

## 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...

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?

l888l888l888 said:
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?

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?

l888l888l888 said:
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

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

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

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

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...

OHHH. ok thanks!

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?

l888l888l888 said:
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.

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.

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?

You just need to modify the first bit:

l888l888l888 said:
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.

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.

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

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

micromass said:
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.

## 1. How is continuity defined for a function?

Continuity for a function is defined as the property of a function where the value of the function at a given input is very close to the value of the function at a nearby input.

## 2. Can a product of two continuous functions be discontinuous?

No, the product of two continuous functions will always result in a continuous function.

## 3. How do you prove that the product of two continuous functions is continuous?

To prove that the product of two continuous functions is continuous, we use the epsilon-delta definition of continuity. This involves showing that for any small value of epsilon, we can find a corresponding delta such that the difference between the product of the two functions at any two points within delta is less than epsilon.

## 4. Is the converse of the statement true? That is, if the product of two functions is continuous, are the individual functions also continuous?

No, the converse of the statement is not always true. It is possible for the product of two functions to be continuous, while one or both of the individual functions may not be continuous.

## 5. Can the product of two continuous functions be discontinuous at a specific point?

Yes, it is possible for the product of two continuous functions to be discontinuous at a specific point. This can occur if one of the functions is not defined at that point, or if there is a discontinuity in one of the functions at that point.

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