Results 1 to 7 of 7

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

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    12

    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?
    Last edited by anon2194; November 17th 2009 at 05:16 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by anon2194 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    12
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by anon2194 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Nov 2009
    Posts
    12
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by anon2194 View Post
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Nov 2009
    Posts
    12
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. prove that every continuous function is separately continuous
    Posted in the Differential Geometry Forum
    Replies: 6
    Last Post: November 23rd 2011, 04:57 AM
  2. Replies: 3
    Last Post: April 18th 2011, 09:19 AM
  3. Replies: 3
    Last Post: April 18th 2011, 08:24 AM
  4. Prove that h is not continuous at 0
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2009, 04:44 PM
  5. prove continuous
    Posted in the Calculus Forum
    Replies: 1
    Last Post: August 12th 2008, 07:44 AM

Search Tags


/mathhelpforum @mathhelpforum