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Assume $f(x)$ is convex in $[0,\infty)$, Prove

$F(x)=\frac{1}{x}\int_0^xf(t)dt$ is convex

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    What did you try? (And please avoid giving orders.)2012-03-14
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    @DidierPiau I don't know where to start2012-03-14
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    What characterizations of convex functions do you know? The more you know, the better...2012-03-14
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    @DidierPiau I know some basic characterizations about convex function.$f(\lambda x_1+(1-\lambda x_2))\leq \lambda f(x_1)+1-\lambda f(x_2)$,and in this case f(x) is continuous in $(0,\infty)$2012-03-14
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    Come! Surely you know some others...2012-03-14
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    @DidierPiau I know I should prove $F(\lambda x_1+(1-\lambda x_2))\leq \lambda F(x_1)+(1-\lambda) F(x_2)$,but I can't handle it.2012-03-14
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    Hint: second derivative.2012-03-14
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    @DidierPiau I don't think it can't be differentiate twice.If it does, I just to prove $F^{''}(x)>0$2012-03-14
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    Of course it cannot *in the general case*, but assuming it can may help you see what is going on.2012-03-14
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    @DidierPiau Can you give me more hints2012-03-14

2 Answers 2

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A hint: Substitute $t:=\tau\, x$ $\ (0\leq\tau\leq 1)$ and look at the resulting integral.

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    Yes,it truly works.But ow can you get out this idea2012-03-14
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    @Gingerjin: Whenever you see ${1\over x}\int_0^x\ldots dt$ you should give this a try.2012-03-14
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0. As alluded to in the comments, there exists several characterizations of convex functions and each of them can prove useful in a given context: one can mention usual convexity inequalities, nondecreasing difference ratios, nondecreasing left- and right-derivatives, nonnnegative second derivatives (when these exist), the graph being above its tangents, some duality à la Fenchel-Legendre, etc.

It happens that a simple change of variables allows to answer the specific question asked here rather shortly. Nevertheless we now explain another method, whose main advantage is that it allows to present some useful tools in this context.

To wit, a characterization of convexity is as follows: the function $f:[0,+\infty)\to\mathbb R$ is convex on $[0,+\infty)$ if and only if there exists a nondecreasing locally integrable function $g:(0,+\infty)\to\mathbb R$ and some finite $c$ such that, for every $x\geqslant0$, $$ f(x)=c+\int_0^xg(y)\mathrm dy. $$ (To be rigorous here, one can only be sure that $f(0)\geqslant c$, but to ask that $f$ is continuous at $0$ does not change its convexity.)

1. In our context, assuming that $f$ is convex and written as above, one gets $$ F(x)=\frac1x\int_0^xf(y)\mathrm dy=c+\frac1x\int_0^x\int_0^yg(z)\mathrm dz\mathrm dy=c+\frac1x\int_0^xg(z)(x-z)\mathrm dz, $$ and one seeks a nondecreasing locally integrable function $h$ such that $F=c+\int\limits_0^{\cdot} h$.

2. To find $h$, introduce the function $$ k=\int\limits_0^{\cdot}yg(y)\mathrm dy. $$ An integration by parts yields $$ F(x)=c+\int_0^xk'(z)\frac{x-z}{xz}\mathrm dz=c+\left[\frac{x-z}{xz}k(z)\right]_ 0^x+\int_0^xk(z)\frac{\mathrm dz}{z^2}. $$ Since $k(z)=o(z)$ when $z\to0$, the second term on the RHS is zero and $F$ has the desired form, with $$ h(x)=\frac1{x^2}k(x)=\frac1{x^2}\int_0^xyg(y)\mathrm dy. $$ 3. To check whether $h$ is nondecreasing, several methods are available here, let us explain one which is completely elementary. For every $0\lt x\leqslant y$, $$ y^2h(y)=\int_0^xzg(z)\mathrm dz+\int_x^yzg(z)\mathrm dz=x^2h(x)+\int_x^yzg(z)\mathrm dz, $$ and $g(z)\geqslant g(x)$ for every $z\geqslant x$, hence $$ 2y^2h(y)\geqslant 2x^2h(x)+g(x)\int_x^y2z\mathrm dz=2x^2h(x)+g(x)(y^2-x^2). $$ Now, $g(z)\leqslant g(x)$ for every $z\leqslant x$ hence $$ 2x^2h(x)\leqslant g(x)\int_0^x2z\mathrm dz=g(x)x^2,\quad\text{that is,}\quad g(x)\geqslant2h(x). $$ Using this in the previous inequality, one gets $$ 2y^2h(y)\geqslant 2x^2h(x)+2h(x)(y^2-x^2)=2y^2h(x), $$ hence $h(y)\geqslant h(x)$ and the proof is complete.

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    You say $f(x)=c+\int_0^xg(y)\mathrm dy$, Is it true when f(x) is not derivable.2012-03-14
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    The second queation is you introduce the function $k=\int\limits_0^{\cdot}yg(y)\mathrm dy$, but is k(z) derivable since g(y) may not be continuous?Please explain it to me,thanks.2012-03-14
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    Yes, this holds also if $f$ is not differentiable (try $f_u(x)=(x-u)^+$). By the way, if $g$ is not continuous, $\int_0^{\cdot}g$ might not be differentiable. // No, $k$ is not differentiable everywhere and $k'$ here is either the left- or the right-derivative of $k$ (as you know, these coincide except on at most a countable subset hence this makes no difference for the integration by parts used).2012-03-14
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    Brilliant.I have a question,and I need your answer. As you can see, I can do some problems rather easy.But when I come across some problems rather complex or I'm not familiar with,i.e.this problem,I always have no idea. Can you give me some suggestions about this?2012-03-14