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Proving continuity of f(x,y) = g(x)p(y)

  1. Jun 22, 2011 #1
    I know this must be easy, but...

    Say real functions g(x) and p(y) are continuous and f(x,y) = g(x)p(y). How to proof rigorously the continuity of f in a point (x1,y1)?

    In other words, how to obtain l g(x)p(y) - g(x1)p(y1) l < epsilon (for any epsilon).

    I can prove that l g(x)p(y1) - g(x1)p(y) l < any epsilon, but I cant see how to go from here to there. I am trying all variations of the triangular inequality, to no avail.

    Thanks for your help.
  2. jcsd
  3. Jun 22, 2011 #2
    We want to prove
    [tex]\forall \varepsilon>0\quad\exists\delta>0\quad\forall (x,y)\in B_{\delta}(x_0,y_0):f(x,y)\in B_{\varepsilon}(f(x_0,y_0))[/tex]

    Let fix epsilon and we want to find such delta. From definition of ball B, we compute


    Now, I think you are able to complete this proof. Use the continuity of g and p.
  4. Jun 22, 2011 #3
    Hi Castilla! :smile:

    [tex]|g(x)p(y)-g(x_1)p(y_1)|=|(g(x)p(y)-g(x_1)p(y))+(g(x_1)p(y)-g(x_1)p(y_1))|\leq |g(x)-g(x_1)||p(y)|+|g(x_1)||p(y)-p(y_1)|[/tex]

    I'll let you continue from there.
  5. Jun 22, 2011 #4
    Thank you, stanley and micromass. You enlightened my mind.
  6. Jun 22, 2011 #5
    Just one question. Looking at both answers, basically the same, it seems that I need that the functions g and p to be bounded.

    So I would need not only that g and p are continuous, but also that they're continuous in a compact set?

    thanks again!
  7. Jun 22, 2011 #6


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    You don't need that, but good catch!

    You have

    |g(x)p(y)-g(x_0)p(y_0)| = |(g(x)-g(x_0))(p(y)-p(y_0)) +p(y_0)(g(x)-g(x_0)) + g(x_0)(p(y)-p(y_0))| <= |g(x)-g(x_0)||p(y)-p(y_0)| +|p(y_0)||g(x)-g(x_0)| + |g(x_0)|p(y)-p(y_0)|
    Last edited: Jun 22, 2011
  8. Jun 22, 2011 #7
    Excellent explanation. Thank you.
  9. Jun 22, 2011 #8


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    Gold Member

    You could have used the fact multiplication is continuous....
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