Explore BrainMass

Explore BrainMass

    Set of functions with a continuous derivative

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    A). Let M be the set of functions defined on [0,1] that have a continuous derivative there ( one-sided derivatives at the endpoints).
    Let p(x,y) = max_[0,1]|x'(t) - y'(t)|.

    1).Show that ( M,p) fails to be a metric space.

    2). Let p(x,y) = |x(0) - y(0)| + max_[0,1]|x'(t) - y'(t)|. Is (M,p) now a metric space?

    Please justify all your answers..I want proofs here not a yes or no answers.

    -----------------------------------------------------------

    B). Let M be the set of continuous functions on [0,1] and define
    p(x,y) = integral from 0 to 1 of |x(t) - y(t)|dt. Does this define a metric space? ( Also a proof here please for the yes or no answer).

    © BrainMass Inc. brainmass.com December 24, 2021, 5:27 pm ad1c9bdddf
    https://brainmass.com/math/derivatives/set-functions-continuous-derivative-48761

    Solution Preview

    A.1
    By definition, a metric d(x,y) must satisfy condition d(x, y) = 0 if and only if x = y.

    It fails for p(x,y) = max_[0,1]|x'(t) - y'(t)|, because two functions can differ by a constant but still have p(x,y) = 0.

    Therefore (M,p) is not a metric space

    A.2
    The definition of metric contains four conditions:
    d(x, y) ≥ 0 (non-negativity) ...

    $2.49

    ADVERTISEMENT