I'm thinking about this and invite you to think too:
Suppose f:R → R is convex and of class C². Show that ∫(0 to 2π) f(x) cos(x) dx ≥ 0
Thank you
Update:We say f is convex on a set S if, for every x1 and x2 in S and every λ in [0, 1], we have
f[λ x1 + (1 - λ) x2)] ≤ λ f(x1) + (1 - λ)f (x2)
That is, the graph of f on [x1, x2] lies below the straight line joining x1 and x2
Update 3:We also see we'll have equality if and only if f(x + π) = f(x) almost everywhere on [0, π]. Something like f being almost periodic with period π if this concept of "almost periodic" makes sense.
Update 5:To Cave: Well, f(x) = ax + b does not satisfy f(x + π) = f(x).
To have equality, I don't think we need to have f(x + π) = f(x) for all x in [0, π], because a non negative function doesn't need to be identrically 0 on the whole interval to have a 0 integral there. It suffices to be 0 almost erywhere on the interval, that is, on a subset of measure 0. OK?
Answers & Comments
Verified answer
Scythian's function f(x) = 1 / (1 + (x-π)²) isn't convex on (0,2π).
I think one fairly straightforward approach would be to take two cases, the first of which is easily disposed of:
i) f monotone
ii) f attains a minimum at (c,f(c)).
(It might be easier to split case i into the separate cases increasing and decreasing, but since cosx is symmetric on the interval, it seems unnecessary.)
If I have more time, I may come back and address this more explicitly; however, we just had a baby today, so I'm kinda doubting it!
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** CLARIFICATION: **
While Steiner's definition is correct, it's also a bit more abstract than necessary in this context.
A convex C² function on the reals just means "concave-up" in the first-semester calculus sense, i.e. d²f/dx² ≥ 0.
In particular, "concave-down" is NOT convex!
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** UPDATE: **
I think Cave has exactly the right approach; unfortunately, it suffers a bit in presentation, so I thought I'd dress it up a bit.
[NOTE: Read carefully; it's very difficult in this font to distinguish f, f', and f'' -- that is, f, df/dx, and d²f/dx².]
Integrating by parts, we have that
∫ f(x) cos(x) dx = f(x)·sinx - ∫ f'(x) sin(x) dx
= - ∫ f'(x) sin(x) dx,
since sin(x) = 0 at both endpoints.
Now, since sin(x) = sin(π-x), we have that
∫ f(x) cos(x) dx = - ∫ f'(x) sin(x) dx
= - ∫ [x=0,π] f'(x) sin(x) dx - ∫ [x=π,2π] f'(x) sin(π-x) dx
= - ∫ [x=0,π] f'(x) sin(x) dx - ∫ [x=0,π] f'(x+π) sin(-x) dx
= - ∫ [x=0,π] f'(x) sin(x) dx + ∫ [x=0,π] f'(x+π) sin(x) dx
= ∫ [x=0,π] ( f'(x+π) - f'(x) )·sin(x) dx.
Finally, since f is convex (i.e. concave-up), f' is increasing, and so f'(x+π) ≥ f'(x) for any x in [0,π].
Therefore we have that ( f'(x+π) - f'(x) )·sin(x) ≥ 0 on [0,π], and so the proof is complete.
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Well, for f(x) = 1-(x-Ï)², it isn't true, so I'll have to find out more about convex functions of "class C² ". I'll get back to you.
All right, I guess you mean "upward convex" functions which is positive from x = 0 to 2Ï. Any straight line f(x) = ax+b gets you 0. Any "upward convex" function can be seen as a downward convex function where f(0) = f(2Ï) = 0 subtracted from any straight line. Since all such downward concave functions result in a negative, the conclusion follows. More on this later. Intuitively speaking, the negative part of Cos(x) would contribute more to the definite integral than the positive parts.
jeredwm, forgive my ignorance about "convex functions", this is actually a new term for me, as I was only aware of convex sets before.
Taking the integral once by parts we get I = integral from 0 to 2*Pi of (-f'(x) * sin(x) dx) because sin(2*Pi) = sin(0) = 0.
f(x) is convex and of class C^2, so f''(x) >= 0 on [0; 2*Pi], therefore f'(x) is non-decreasing on [0; 2*Pi].
Sin(x) on [0; 2*Pi] consists of 2 identic arcs, on (0; Pi) it is positive and on (Pi; 2*Pi) it is negative. As f'(x) is non-decreasing then integral of f'(x) * |sin(x)| dx is not greater on [0; Pi] than on [Pi; 2*Pi], that is why the full integral is not above zero and thus minus that integral is not below zero.
Speaking more strictly, sin(x+Pi) = -sin(x), so letting g(x) = f'(x) we have integral from 0 to 2*Pi of g(x) * sin(x) dx = integral from 0 to Pi of g(x) * sin(x) dx + integral from Pi to 2*Pi of g(x) * sin(x) dx = [subbing t = x - Pi in the second integral] = integral from 0 to Pi of g(x) * sin(x) dx + integral from 0 to Pi of g(Pi+t) * (-sin(t)) dt = integral from 0 to Pi of (g(x) - g(Pi+x)) * sin(x) dx <=0 because sin(x) >= 0 and g(x) <= g(Pi+x) because g(x) is non-decreasing.
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jeredwm,
thank you for your post. With all the respect I don't see much difference in your proof and mine. Though I can't judge if it is your own or "cleaned" mine.
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Almost, we have equality if and only if f'(x) = f'(x+Pi) for all x in [0;Pi]. This is equivalent to f(x+Pi) = f(x) +C, C is a constant. For example, for all linear functions f(x) = ax + b will do.
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It depends on which integral is being used. If Lebesgue one that yes, true, for almost all x. My note was not about it but about those ` ' ` signs near f(x): the condition uses df/dx, not f(x) itself.