# Math Help - If f and g are continuous, prove that fg is continuous

1. ## If f and g are continuous, prove that fg is continuous

$f,g : A \rightarrow R$ are continuous on A.

We are given that:
$\forall c \in A, \forall \epsilon > 0, \exists d_{f}(\epsilon, c): \forall x \in A, 0 < |x - c| < d_{f} \rightarrow |f(x) - f(c)| < \epsilon$
$\forall c \in A, \forall \epsilon > 0, \exists d_{g}(\epsilon, c): \forall x \in A, 0 < |x - c| < d_{g} \rightarrow |g(x) - g(c)| < \epsilon$

$\forall x,c \in A$,
$|fg(x)-fg(c)| = |fg(x)-f(x)g(c)+f(x)g(c)-fg(c)|$
$|fg(x)-fg(c)| \le |f(x)|*|g(x)-g(c)|+|g(c)|*|f(x)-f(c)|$

Let $M(x,c)=$max{ $|f(x)|, |g(c)|$}

Then $|fg(x)-fg(c)| \le M(x,c)*(2\epsilon)$

i.e. $\forall c,x \in A, \forall \epsilon_{1}:=\frac{\epsilon}{2M(x,c)}>0, \exists \delta_{fg}=$min{ $d_f(\epsilon,c),d_g(\epsilon,c)$} $:0<|x-c|<\delta(\epsilon,c) \rightarrow |fg(x)-fg(c)|<\epsilon_{1}$

Then fg is continuous on A.

This is my proof for the stated theorem. Is there any way I can improve it?

2. Originally Posted by anon2194
$f,g : A \rightarrow R$ are continuous on A.

We are given that:
$\forall c \in A, \forall \epsilon > 0, \exists d_{f}(\epsilon, c): \forall x \in A, 0 < |x - c| < d_{f} \rightarrow |f(x) - f(c)| < \epsilon$
$\forall c \in A, \forall \epsilon > 0, \exists d_{g}(\epsilon, c): \forall x \in A, 0 < |x - c| < d_{g} \rightarrow |g(x) - g(c)| < \epsilon$

$\forall x,c \in A$,
$|fg(x)-fg(c)| = |fg(x)-f(x)g(c)+f(x)g(c)-fg(c)|$
$|fg(x)-fg(c)| \le |f(x)|*|g(x)-g(c)|+|g(c)|*|f(x)-f(c)|$

Let $M(x,c)=$max{ $f(x), g(c)$}

Then $|fg(x)-fg(c)| \le M(x,c)*(2\epsilon)$

i.e. $\forall c,x \in A, \forall \epsilon_{1}:=\frac{\epsilon}{2M(x,c)}>0, \exists \delta_{fg}=$min{ $d_f(\epsilon,c),d_g(\epsilon,c)$} $:0<|x-c|<\delta(\epsilon,c) \rightarrow |fg(x)-fg(c)|<\epsilon_{1}$

Then fg is continuous on A.

This is my proof for the stated theorem. Is there any way I can improve it?

Perhaps just to explain briefly why $M(x,c)$ exists...(continuous function in closed bounded interval...etc.)

Tonio

3. I thought M(x,c) exists just because f and g are defined at x and c. Also, I don't think A has to be closed, bounded, or even an interval for this proof.

And I realized that M(x,c) could be zero... *edit* So I went back and changed the definition of M(x,c). I added the absolute values. Then M(x,c) = 0 iff f(x)=g(c)=0. In that case |f(x)g(x) - f(c)g(c)| = |0 - 0| = 0, and we can just let e_{1} be any in the case of M(x,c) = 0.

4. Originally Posted by anon2194
I thought M(x,c) exists just because f and g are defined at x and c. Also, I don't think A has to be closed, bounded, or even an interval for this proof.

Oh, but it is VERY important! Otherwise how can you show M exists? For example, take $f(x)=\sqrt{x}\,\,in\,\,(0,1)$. This function isn't bounded in this interval and remember: you defined M to be what it is but x can be ANYTHING...

And I realized that M(x,c) could be zero...
Zero is fine: who cares?

I thought you intended to prove that if f,g are continuous at some point then their product is continuous at that point. If you intended something else then your proof may well be false.

Tonio

5. Still not following. Given an x and a c, f(x) and g(c) are defined and in R. Then either M = g(c) if f(x) <= g(c), or M = f(x) if g(c) <= f(x). M always exists, so the type of set A is is just restricted by the fact that f and g are continuous on it.

And I am trying to prove that if f,g are continuous at some point c then their product is continuous at that point c. And c is any point in A.

6. Originally Posted by anon2194
Still not following. Given an x and a c, f(x) and g(c) are defined and in R. Then either M = g(c) if f(x) <= g(c), or M = f(x) if g(c) <= f(x). M always exists, so the type of set A is is just restricted by the fact that f and g are continuous on it.

And I am trying to prove that if f,g are continuous at some point c then their product is continuous at that point c. And c is any point in A.

I understand all what you wrote but you are no taking one definite, fixed x but ALL the x's s.t. $|x-c|<\delta$ . The point that is, or should be, fixed is c, the point at which you need to prove continuity! How can you prove that f(x) is bounded no matter what x you choose in $(c-\delta,c+\delta)$?? you need to use, imo, Weierstrass' Theorem on boundeness of functions in closed intervals.

Tonio

7. Ok, so if A is closed and bounded, and f is continuous on A, then f is bounded on A.

Then there is an M such that for all x,c in A, f(x)g(c) < M. Choose that as M, so M does not depend on c or x.

Then move $\forall x \in A$ into its proper position after $\exists \delta ...$. Thanks man, that helped allot (assuming what I just said fixes everything). I had a feeling that moving $\forall x \in A$ screwed up something.