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Real analysis - show convex functions are left & right differentiable

  1. Nov 22, 2007 #1

    quasar987

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    [SOLVED] Real analysis - show convex functions are left & right differentiable

    1. The problem statement, all variables and given/known data

    Let f:R-->R be convex. Show f admits in every point a left derivative and a right derivative.

    2. Relevant equations

    A function f:R-->R is convex if x1 < x < x2 implies

    [tex]f(x)\leq \frac{x_2-x}{x_2-x_1}f(x_1)+\frac{x-x_1}{x_2-x_1}f(x_2)[/tex]

    Or equivalently, if whatever x, y, and [itex]\lambda[/itex] in [0,1],

    [tex]f(\lambda x + (1-\lambda)y\leq \lambda f(x) + (1-\lambda)f(y)[/tex]


    By left derivative at x0, we mean the limit

    [tex]D_lf(x_0)\lim_{x\rightarrow x_0^-}\frac{f(x)-f(x_0)}{x-x_0}[/tex]

    and by right derivative at x0, we mean the limit

    [tex]D_rf(x_0)\lim_{x\rightarrow x_0^+}\frac{f(x)-f(x_0)}{x-x_0}[/tex]

    3. The attempt at a solution

    Let's stick to the left derivative.

    I know convex functions are Lip****z, so the differential quotient is bounded.

    I have also proven in an earlier exercise that the differential quotient is increasing as x increases:

    "If x1 < x < x2, then [tex]\frac{f(x)-f(x_1)}{x-x_1}\leq \frac{f(x_2)-f(x_1)}{x_2-x_1} \leq\frac{f(x_2)-f(x)}{x_2-x}[/tex]"

    But this does not give the conclusion because I must show the differential quotient converges for any sequence, monotonous or not, converging to x0.

    Can we show limsup=liminf? Can we show it is Cauchy? I don't see how.
     
    Last edited: Nov 22, 2007
  2. jcsd
  3. Nov 23, 2007 #2

    morphism

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    Keep in mind that you're taking the limit as x approaches x_0 from the left. So out of the sequences converging to x_0, we need only consider those whose tails increase to x_0.
     
  4. Nov 23, 2007 #3

    quasar987

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    how come? I thought limit from the left only meant that we only consider sequences whose points are lesser than x_0
     
  5. Nov 23, 2007 #4

    morphism

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    I worded that very badly!

    What I was trying to get across is that if L = lim(x->a-) g(x) exists, and g(x) is increasing, then this limit is going to be sup{g(x) : x < a} (by uniqueness of limits).
     
  6. Nov 23, 2007 #5

    quasar987

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    this allows us to conclude that the limit of the differential quotient exists?
     
  7. Nov 23, 2007 #6

    morphism

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    Let g be the differential quotient. Why does sup{g(x) : x < x_0} exist?
     
  8. Nov 23, 2007 #7

    quasar987

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    I see your point! Now I can try to show directly that the limit is sup{g(x) : x < x_0}.

    And this is easy! you rock :D
     
    Last edited: Nov 23, 2007
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