1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook