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A question regarding convex function.

  1. Jun 26, 2006 #1
    let f:(a,b)-> a differentiable function, f is a convex function iff for every x in (a,b) the tangent line to f's graph in x is below f.

    i tried it this way:
    suppose f is a convex function, then for every 0<d<1 and every s,t in (a,b), f(dt+(1-d)s)<=df(t)+(1-d)f(s), now the tangent line in lets say x0 is
    g(x)=f'(x0)(x-x0)+g(x0)
    if s>t and t<x<s then g(t)=f'(x0)(t-x0)+g(x0) g(s)=f'(x0)(s-x0)+g(x0)
    f'(x0)=[g(t)-g(s)]/(t-s)
    i need to prove that g(x)<=f(x)
    now if x=dt+(1-d)s then g(dt+(1-d)s)=f'(x0)(dt+(1-d)s-x0)+g(x0)

    how do i proceed from here?, thanks.
     
  2. jcsd
  3. Jun 26, 2006 #2

    0rthodontist

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    Well, the first thing is that g(x) = f'(x0)(x-x0)+f(x0). Your equation for g is correct but hard to work with.

    I'm not sure how to approach this. If you could show that f is _locally_ greater than its tangent line, then it is not hard to show that it is globally greater.
     
  4. Jun 26, 2006 #3

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    Does it help to know that a line is just convex, in the sense that the inequality you mentioned is actually an equality for all d, t, and s?
     
  5. Jun 27, 2006 #4
    more questions regarding convex function:
    1) let f:[a,b]->R be a convex function, prove that if there exists c in [a,b] such f(a)=f(b)=f(c) then f is constant.
    2) let g:(0,infinity)->R be a convex function ,
    lim g(x)<infinity
    x->infinity
    prove that g(x) monotonic decreasing.

    what I did is as follows:
    1) for every 0<d<1 f(c)=df(a)+(1-d)f(b)>=f(da+(1-d)b)
    let c=(1-d)a+db then f(c)=f((1-d)a+db)<=(1-d)f(a)+df(b)=f(c)
    f(da+(1-d)b)<=df(a)+(1-d)f(b)=f(c)=f(1-d)a+db)<=(1-d)f(a)+df(b)=f(c)
    how do conclude from this that fis constant.
    2) from what is given we can deduce that: g(x)<=M when M is the supremum of g(x), now we also have for x1>x2 and 0<d<1 g(dx2+(1-d)x1)<=dg(x2)+(1-d)g(x1) if we let M=dg(x2)+(1-d)g(x1) then g(x)<=dg(x2)+(1-d)g(x1)
    if x=x1 then g(x1)<=g(x2) but also when x=x2 g(x2)<=g(x1)
    so I reackon my approach is flawed so how do I fix it?
    thanks in advance.
     
  6. Jun 27, 2006 #5

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    For the first one, assume some point d, say, between a and c, has f(d) not equal to f(a). Depending on whether it is greater or less, you can find points between which the function contradicts the convex assumption. (try drawing it)

    For the second, assume there is some x,y with x<y such that f(x)<f(y). Apply the mean value theorem to show there is some point between x and y with positive slope, and then use the results of the question in your first post to show the function cannot have a finite limit at infinity.
     
  7. Jun 29, 2006 #6
    so is this proof correct?

    let y>x g(y)>=g(x) let c in (x,y) g'(c)=[g(y)-g(x)]/(y-x)>=0
    let f(x) be the tangent line to g(x) in the point (c,g(c))
    f(x)=g'(c)(x-c)+g(c)
    because g(x) is convex function, g(x)>=g'(c)(x-c)+g(c)
    therefore g(x) is not bounded above, and thus doesnt converge as x approaches infinity.
     
  8. Jun 29, 2006 #7

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    That looks about right, except it should be g(y)>g(x), not g(y)>=g(x). And maybe add one more step before the end. "Not bounded above" does not directly follow from "bounded below by f(x)," expecially if f(x) is constant as your proof curently allows (see my first point).
     
    Last edited: Jun 29, 2006
  9. Jun 29, 2006 #8
    but im not given in my second question that g is constant.
    so what should i add?, i thought that it suffices to say that g(x) is not bounded above to show that as x apparoaches infinity g(x) diverges.
     
  10. Jun 29, 2006 #9

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    A monotonically decreasing function satisfies f(x)>=f(y) whenever x<y. You are assuming that the function is not monotonically decreasing, that is, that there is some x<y with f(x)<f(y), and then showing that if the function is also convex, this implies it must blow up at infinity. Do you see why this gives you what you want?

    And you haven't shown that it's not bounded above. To do this you must show that for any number b, there is some a for which x>a implies f(x)>b. Bounding something below by a function f(x) does not say it isn't bounded above unless you first prove something about f(x). Maybe I'm being picky, but I think you should show this step.
     
    Last edited: Jun 29, 2006
  11. Jun 29, 2006 #10
    so if i put x=c then f(c)=g(c), and for every M=g(c) when x>c then g(x)>=M will that suffice?

    btw, i know this epsilon delta definition, and i understand, that you're employing an ad absurdum proof but what i didn't understand is why did you say:"expecially if f(x) is constant as your proof curently allows " why my proof allows g (or as you say f(x)) to be constant?

    by the way the definition i know of monotonic decreasing, is if x>y then f(y)>f(x), this is why i assumed that x>y and f(x)>=f(y) (which is the negation of f(y)>f(x)).
     
  12. Jun 29, 2006 #11

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    I have no idea what you're saying here.

    I'm referring to the following:

    Here you use g(x) to refer to the original function, and f(x) to refer to the tangent line. Since g'(c) may be zero in this argument, f(x) may be constant, in which case g(x) may in fact be bounded above, even if it is bounded below by f(x).

    "Monotonically decreasing" and "strictly monotonically decreasing" are different concepts. The authors of this question must mean the former, not the latter (which you are describing here), because the function f(x)=0 is (a) convex, (b) has a finite limit as x goes to infinity, and (c) not strictly monotonically decreasing , thus providing a counterexample to the statement you want to prove. However, f(x)=0 is monotonically decreasing, which is to say it satisfies the conditions I spelled out in my last post.
     
    Last edited: Jun 29, 2006
  13. Jun 30, 2006 #12
     
  14. Jun 30, 2006 #13

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    From what you say in the second paragraph, you know that isn't correct, because you've only shown what you need to for one number, f(c). Maybe I've overcomplicated things. I'll explain what you need to do in english and let you translate it to math. You have a function that is bounded below by a line which has positive slope. It is easy to prove that the line is not bounded above. Thus the function cannot be bounded above.
     
  15. Jun 30, 2006 #14
    tell me if this is right:
    for every real number b, such that there exists x, which x>c then f(x)>f(c) (this we can conclude from our assumption), if we let f(c)>b, then we have that f(x) diverges, and thus not bounded above.
     
  16. Jun 30, 2006 #15

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    What is c here? Is it the point where the function has positive slope, as it was before? If so, remember: you only know it has positive slope at one point, so you can't just keep increasing f(c) so that it is greater than any given b. f(c) is fixed.

    All you need to do is prove that a line is not bounded above. For example, the line f(x)=2x is not bounded above because for any a we might pick, we can find a number b such that x>b implies f(x)>a, namely, by taking b>=a/2.

    Once you prove the tangent line, which you originally called f(x), is not bounded above, a simple argument shows that any function g(x) such that g(x)>=f(x) for all x (which your original function has been proven to satisfy) must also be unbounded above. You should not need to reference the point x=c or f(c) again.

    And please make an earnest attempt to solve the problem. You have more than enough information. I'll only help you again if you have a question that you can't answer even after trying very hard to figure it out yourself. That's the only way you'll learn anything.
     
    Last edited: Jun 30, 2006
  17. Jul 1, 2006 #16
    ok, if we let x>b then for every a which b>=(a/g'(c)+c-g(c)/g'(c)) we have that f(x)>f(b)>=f(a/g'(c)+c-g(c)/g'(c))=a.

    if this is right, it's quite messy.
     
  18. Jul 1, 2006 #17

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    That's the idea. It might actually be easier and neater to first prove a more general result, like that all lines of positive slope are not bounded above, or that if one function is not bounded above, any function differing from it by an additive constant is also not bounded above. Then you can apply this result to your example. But what you have will suffice.
     
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