Problem With Proof?

sponsoredwalk

Questions are offered when you see a [☺]:

The following theorem, "the 'basic lemma' of the calculus of variations",
on page 1 of this book is the following:

"If f is a continuous function in [a,b], with a < b, s.t. ∫ₐᵇη(x)f(x)dx = 0 for
an arbitrary function η continuous in [a,b] subject to 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 [1] technically does that
not just prove the theorem for this function alone, not for any "arbitrary"
function? I see how this easily extends to positive functions though &
obviously negative ones too.

But [2] 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] to satisfy ∫ₐᵇη(x)f(x)dx = 0.
My concern here is that if we were to trust the method of proof used in
the book for η = 0 then we'd conclude f = 0 when it need not be.
If, on [a,b], η(x) = 0 & f(x) = 2x then ∫ₐᵇη(x)f(x)dx = ∫ₐᵇ0·2xdx = 0 but 2x ≠ 0.

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 [3] 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 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 that ∫ₐᵇη(x)f(x)dx = 0.

But [4] this brings in to question another concern, f could be non-zero at
every other point on [a,b] except at least at the non-zero value η(cᵢ)
we're forced to assume exists, 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 + ... = 0

Here you'd satisfy the hypothesis by having the sum equal to zero but
the conclusion doesn't follow! f(x) = 0 only at certain parts, very devious!
The flaw lies in the inclusion of the phrase "arbitrary function" as far as I
can see, I think it should be "arbitrary non-zero function". Thoughts?

Fremen

[1]: I'm not sure why they would do this. Maybe they're trying to make the steps of the proof clear - I'd guess they're integrating by parts, are they?

If you proved the result for all polynomials (x-x1)(x-x2)...(x-xn), then the theorem would hold by linearity of integration and the Stone-Weierstrass theorem. Maybe they're just showing how to do the inttegration in the simplest case.

[2]:

There's an analogy with vector spaces here (actually, it's more than an analogy, what we're looking at is a vector space modulo some technicalities). One way to prove that a vector V is 0 is to show that V.X = 0 for all X (this is the dot product, which I'm not bothered writing in latex).

It's not sufficient to pick X=0, take the dot product V.0 = 0 and conclude that V=0. It has to hold for all X at the same time.

Same thing with functions.

Does this answer your concern about [3] and [4]?

Michael Collins

OK I'll admit I haven't gone through each of your questions in detail but I wonder could this just be a problem with notation here?

The book says: Let [latex] f(x) [/latex] be a continuous function such that

[latex] \displaystyle \int_a^b \eta(x) f(x) \hbox{d}x = 0[/latex]

for any arbitrary function [latex] \eta(x) [/latex].

i.e. this must be satisfied for ANY function, not just [latex] \eta(x) = 0 [/latex] but the one they give as well: [latex] \eta(x) = (x-x_1)(x-x_2) [/latex]. So it's a very strong condition.

equivariant

sponsoredwalk wrote: »

Questions are offered when you see a [☺]:

The following theorem, "the 'basic lemma' of the calculus of variations",
on page 1 of this book is the following:

"If f is a continuous function in [a,b], with a < b, s.t. ∫ₐᵇη(x)f(x)dx = 0 for
an arbitrary function η continuous in [a,b] subject to 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 [1] technically does that
not just prove the theorem for this function alone, not for any "arbitrary"
function? I see how this easily extends to positive functions though &
obviously negative ones too.

I have read through their proof and I see no problem. Their strategy is to show that if f is non zero at any point, that will somehow contradict the hypothesis. To arrive at this, they choose a particular function \eta (the one you mention) and show that if f(t) is non zero for some particular t and \eta is chosen correctly (ie the values of x_1 and x_2 are chosen sufficiently close to t) then the hypothesis of the lemma will be contradicted by that particular choice of \eta. But since the hypothesis is assumed to be true for ALL \eta, then our assumption that f(t) is non zero must in fact be false. But t was arbitrary, so we have shown that f is identically zero.

sponsoredwalk

Michael Collins wrote: »

i.e. this must be satisfied for ANY function, not just [latex] \eta(x) = 0 [/latex] but the one they give as well: [latex] \eta(x) = (x-x_1)(x-x_2) [/latex]. So it's a very strong condition.

Yeah that's the point I'm making, the function they've given proves it for
the function η(x) = (x - x₁)(x₂ - x) > 0. Obviously you can reverse this to
show it holds for η(x) = (x₁ - x)(x₂ - x) < 0.I take it that these serve as a
means to show the therem is supposed to hold for any arbitrary positive &
negative function. My problem is both with the case of η(x) = 0 on [a,b] &
further with the case where η(x) = 0 when f(x) ≠ 0 (& vice versa) which is
clearer by forming a Riemann sum (as I did at the end of the post), it's the
ANY η function stipulation I can't justify.

Fremen wrote: »

[1]: I'm not sure why they would do this. Maybe they're trying to make the steps of the proof clear I'd guess they're integrating by parts, are they?

I think they specified this function because of the a < b condition, it is a
clear way to make the function positive by choosing (x - x₁)(x₂ - x), I
think you can justify this because the important step of the proof is to
show that we contradict ∫ₐᵇη(x)f(x)dx = 0 by having a positive functionη
& also a positive f. But I think this might be flawed in the ways I've
mentioned.

Fremen wrote: »

If you proved the result for all polynomials (x-x1)(x-x2)...(x-xn), then the theorem would hold by linearity of integration and the Stone-Weierstrass theorem.

Well ignoring the η = 0 case & the jigsaw case I illustrated in the picture
below I see what you mean, I think the 'proof' I gave in [3] would also do
it unless you have a problem with it.

Fremen wrote: »

[2]:

There's an analogy with vector spaces here (actually, it's more than an analogy, what we're looking at is a vector space modulo some technicalities). One way to prove that a vector V is 0 is to show that V.X = 0 for all X (this is the dot product, which I'm not bothered writing in latex).

It's not sufficient to pick X=0, take the dot product V.0 = 0 and conclude that V=0. It has to hold for all X at the same time.

Same thing with functions.

But here I don't see the flaw with choosing η(x) = 0 when f(x) ≠ 0 &
f(x) = 0 when η(x) ≠ 0 on [a,b], that way you still satisfy all the
conditions of the hypothesis but clearly f(x) ≠ 0.

equivariant wrote: »

But since the hypothesis is assumed to be true for ALL \eta, then our assumption that f(t) is non zero must in fact be false.

Surely assuming the hypothesis is true for ALL η means we're including
those η that have at least one value equal to zero on [a,b], say η(x₃) = 0,
where a < x₃ < b, then:

∑η(xᵢ)f(xᵢ)δxᵢ = η(x₁)·f(x₁)δx₁ + η(x₂)·f(x₂)δx₂ + η(x₃)·f(x₃)δx₃ + ... = η(x₁)·0δx₁ + η(x₂)·0δx₂ + 0·f(x₃)δx₃ + ... = 0

But here we see that f(x₃) could be anything, i.e. f(x) ≠ 0 on [a,b].
Now, surely there are many, many, continuous functions η that exist
on [a,b] that allow for a non-zero continuous function f of the form:



So f(x) = 0 when η(x) ≠ 0 & η(x) = 0 when f(x) ≠ 0.
Here lim ∑η(xᵢ)f(xᵢ)δxᵢ = ∫ₐᵇη(x)f(x)dx = 0 but f(x) ≠ 0.
I just don't see the problem with this, we satisfy the hypothesis but f(x) ≠ 0.

The point is that if I arbitrarily pick η = 0 on all of [a,b] I don't see how
the proof in the book (or even the theorem) is justified & I also don't see
how the theorem holds for functions that obey the kind of thing going on
in the picture, it can't be true for every function η only some η, I interpret
the inclusion of the word "arbitrary" as meaning it doesn't matter what
function η we pick the theorem should still hold, is that incorrect? If not
then what's wrong with these 'counterexamples'?

Michael Collins

The theorem doesn't say: "For any n(x) if the integral is zero, then f(x) = 0 everywhere in that interval".

It says:

If we do this integral for every n(x) we can possibly think of (subject to the continuity constraint), and we find the integral comes out to be zero each time, then f(x) must be identically zero in that interval.*

To prove this it uses contradiction:

1) So we assume the integral does come out to be zero for every n(x) we can think of, but that the function f(x) is not identically zero. (This is the premise which we'll prove to be false by arriving at a contradiction.)

2) Now if we are to prove the premise 1 above, we must show that for every function n(x), the integral of some non-identically zero function f(x) when multiplied by n(x), comes out to be zero.

3) But if we are to disprove this premise, all we need is one counter-example. And they find it: (x-x1)(x-x2).

*Continuity is very much necessary here. If continuity wasn't the case, then this would be false, since the function

f(x) = 1 at some point in a<x<b but f(x) = 0 everywhere else in a<x<b

would integrate to zero for every n(x) (even n(x) = 1), but clearly it's not identically zero in that interval!

sponsoredwalk

Where do the conditions I explained, with reference to the picture, at the
end of my last post fail to satisfy your criteria?

I know what's going on but I believe I've found a counter-example that
satisfies the conditions of the hypothesis (continuity, a < b, η(a) = η(b) = 0)
but still doesn't imply that f(x) = 0 on [a,b]. Could you point out how my
apparent counter-example in the picture fails in some way as I can't!?

Michael Collins

sponsoredwalk wrote: »

Where do the conditions I explained, with reference to the picture, at the
end of my last post fail to satisfy your criteria?

I know what's going on but I believe I've found a counter-example that
satisfies the conditions of the hypothesis (continuity, a < b, η(a) = η(b) = 0)
but still doesn't imply that f(x) = 0 on [a,b]. Could you point out how my
apparent counter-example in the picture fails in some way as I can't!?

Because your counter-example is only one specific n(x), the theorem requires the integral to be zero for every n(x), not just your one, but any you can possibly imagine. They say this by using the phrase "...any arbitrary function n(x)..." but maybe it'd be better stated "...every arbitrary function n(x)...".

Fremen

Using the vector analogy, you're setting

v = (17,21,3,0) and eta = (0,0,0,5), say.

eta . v = 0 for this choice of eta.

However, pick

eta2 = (1,0,0,0), and

eta2.v = 17

so it's not true that

x.v = 0, for all x so we can't conclude that v is 0.

If you want to convert this into an almost-rigorous example, divide the interval [0,1] up into four intervals I1, I2, I3 and I4. Let v(x) take the value 17 on most of the interval I1, 21 on most of I2, 3 on most of I3, and 0 on most of I4. Interpolate smoothly between the values. Similarly with eta and eta2. The integrals will be nearly 0 and nearly 17 respectively. Hope that makes sense to you.

sponsoredwalk

Ah, I get it now. I was slightly misinterpreting it, so while I could find
a function f to make f(x) = 0 when η(x) ≠ 0 & η(x) = 0 when f(x) ≠ 0,
that same function f will not have ∫ₐᵇη(x)f(x)dx = 0 for a different choice of
η. It's like I stopped thinking halfway through a sentence because this
should have been immediately apparent to me!

Fremen

The phrase "for any X" has tripped me up in the past too. My natural interpretation would have been the one you used, but that's not the standard interpretation in maths. Maybe it's an Irish thing, I dunno.