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Basic Lemma

  1. Aug 15, 2011 #1
    The following theorem is called "the 'basic lemma' of the calculus of variations" on page 1
    of this book:

    "If f is a continuous function in [a,b] s.t. ∫abη(x)f(x)dx = 0 for an arbitrary function η
    continuous in [a,b] subject to the condition that η(a) = η(b) = 0 then f(x) = 0 in [a,b]"

    If you read the proof you'll see they go ahead & specify the function η by (x - x₁)(x₂ - x)
    & prove the claim using that, but technically does that not just prove the theorem for this
    function alone, not for any arbitrary function?

    Also, if we arbitrarily choose η to be the zero function s.t. η(x) is zero on [a,b] then f need
    not equal zero on [a,b] for this theorem to hold. Surely I'm missing something?

    Assuming I'm right, we must modify the hypothesis to make η non-zero at least once on
    [a,b] & choose η so that it is non-zero at least once on [a,b], now could it be considered a
    proof by way of contradiction to simply take advantage of the limit of a sum formulation of
    the integral & try to prove it using an arbitrary η:

    Using |∑η(xᵢ)f(xᵢ)δxᵢ - 0| < ε we see that this reduces to|∑η(xᵢ)f(xᵢ)δxᵢ| < ε.
    As we've assumed η can be arbitrary if it's non-zero at least once on [a,b] then the sum
    ∑η(xᵢ)f(xᵢ)δxᵢ will equal at least one definite value as f is assumed to be non-zero on
    [a,b]. But now there exists an ε ≤ |∑η(xᵢ)f(xᵢ)δxᵢ|, contradicting our original assumption.

    But this brings in to question another concern, f could be non-zero at every other
    point on [a,b] except the non-zero value η(cᵢ) we're forced to assume exists as above,
    what I mean is:

    ∑η(xᵢ)f(xᵢ)δxᵢ = η(x₁)f(x₁)δx₁ + η(x₂)f(x₂)δx₂ + ... = 0·f(x₁)δx₁ + 0·f(x₂)δx₂ + ... + η(cᵢ)·0δxᵢ + ...

    Here you'd satisfy the hypothesis by having the sum equal to zero but the conclusion
    doesn't follow! The flaw lies in the inclusion of the phrase "arbitrary function" as far as
    I can see.

    I really feel I must be making a basic, basic, error in my interpretation of this frankly but
    as it stands I just don't see where I'm wrong. Please let me know :cool:
     
  2. jcsd
  3. Aug 16, 2011 #2

    CompuChip

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    Science Advisor
    Homework Helper

    The statement is basically:
    If f is continuous and for all [itex]\eta[/itex] <something holds>, then f is zero.

    The proof is by contradiction: assume that f is non-zero, and prove something in contradiction with the assumption.
    The negation of "f is continuous and for all [itex]\eta[/itex] <something holds>" is "f is not continuous or for some [itex]\eta[/itex] not <something holds>".
    So if we also assume that f is in fact continuous, we derive the contradiction by showing that for some [itex]\eta[/itex], the <something> doesn't hold. There may be more than one such [itex]\eta[/itex], but the proof just gives you one (and shows that it will do).
     
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