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Homework Help: One more H.W. question

  1. Jun 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Let f and g be functions from R to R. For the sum and product of f and g, determine which statements below are true. If true, provide a proof; if false, provide a counterexample.

    a) If f and g are bounded, then f + g is bounded
    b) If f and g are founded, then fg is bounded
    c) If f+g is bounded, then f and g are bounded
    d) If fg is bounded, then f and g are bounded

    2. Relevant equations


    3. The attempt at a solution

    "Bounded" just means in the real-numbered set S there is a real number M such that |x|≤M for all x in S.

    So, say F is the max for f and G is the max for G.

    For example, say f(x)=5-x2 and g(x)=6-x2. F=5, S=6.

    f(x) + g(x) = 11-2x2.

    Still bounded, of course. But how do I give proofs of all these? Give me an example or two.
  2. jcsd
  3. Jun 24, 2010 #2
    Exactly as you said. Consider (a). If f is bounded by F and g is bounded by G, that means for all x in R, |f(x)| ≤ F and |g(x)| ≤ G. So, |(f+g)(x)| = |f(x)+g(x)| ≤ ? ≤ M. What is M and what goes in the question mark? That is how such a proof would go.
  4. Jun 24, 2010 #3
    M = G + F

    Not sure about the question mark
  5. Jun 24, 2010 #4
    True, M = G + F, but you cannot just directly conclude |f(x)+g(x)| ≤ G + F. There needs to be an additional step in there, which is the question mark. Remember the definition of F and G.
  6. Jun 25, 2010 #5


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    Science Advisor

    No, if f(x) is bounded then there exist M such that |f(x)|< M. If g(x) is bounded, there exist N such that |g(x)|< N.

    Now, [itex]|f(x)+ g(x)|\le |f(x)|+|g(x)|< [/itex] what?

    [itex]|f(x)g(x)|\le |f(x)||g(x)|< [/itex] what?
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