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Let $f:\mathbb R\to \mathbb R$ be a convex function of class $C^1$. Let us suppose that there exists the limit

$$L:=\lim_{x\to+\infty}\frac{f(x)}{x^2}$$ and that $0<L<+\infty$.

a) Prove that $$0<\liminf_{x\to+\infty}\frac{f'(x)}{x}\leq\limsup_{x\to+\infty}\frac{f'(x)}{x}<+\infty.$$

b) Prove that there exists the limit $$\lim_{x\to+\infty}\frac{f'(x)}{x},$$ and compute it.


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it can help if you write $f(x) = Lx^2+\varepsilon(x)x^2$ where $\lim\limits_{x\to\infty}\varepsilon(x) = 0$. – Ilya Sep 1 '11 at 12:47
Hint: the Mean Value Theorem says that for some $\xi$ so that $y<\xi<x$,$$\frac{f(x)-f(y)}{x^2-y^2}=\frac{f'(\xi)}{2\xi}$$ If that is not enough, I have a complete proof I can post. – robjohn Sep 1 '11 at 16:23
up vote 3 down vote accepted

Pick any $\epsilon>0$, then there is an $x_0$ big enough so that for $x>x_0$, $\left|\frac{f(x)}{x^2}-L\right|<\epsilon^2$. Then, for $x>y>x_0$, $$ \begin{align} \frac{f(x)-f(y)}{x^2-y^2}&\le\frac{(L+\epsilon^2)x^2-(L-\epsilon^2)y^2}{x^2-y^2}\\ &=L+\epsilon^2\frac{x^2+y^2}{x^2-y^2} \end{align} $$ and $$ \begin{align} \frac{f(x)-f(y)}{x^2-y^2}&\ge\frac{(L-\epsilon^2)x^2-(L+\epsilon^2)y^2}{x^2-y^2}\\ &=L-\epsilon^2\frac{x^2+y^2}{x^2-y^2} \end{align} $$ Thus $$ \left|\frac{f(x)-f(y)}{x^2-y^2}-L\right|\le\epsilon^2\frac{x^2+y^2}{x^2-y^2}\tag{1} $$ Choose $x>y>z>x_0$ so that $\frac{x^2-y^2}{x^2+y^2}=\epsilon$ and $\frac{y^2-z^2}{y^2+z^2}=\epsilon$. By the Mean Value Theorem, for some $\xi$ and $\eta$ so that $x>\xi>y>\eta>z$, we have $$ \frac{f'(\xi)}{2\xi}=\frac{f(x)-f(y)}{x^2-y^2}\text{ and }\frac{f'(\eta)}{2\eta}=\frac{f(y)-f(z)}{y^2-z^2}\tag{2} $$ Using $(1)$ and $(2)$, $$ \left|\frac{f'(\xi)}{2\xi}-L\right|<\epsilon\text{ and }\left|\frac{f'(\eta)}{2\eta}-L\right|<\epsilon\tag{3} $$ Note that with our choice of $x$, $y$, and $z$, we have $$ 1\le\frac{\xi}{y}\le\frac{x}{y}=\sqrt{\frac{1+\epsilon}{1-\epsilon}}\text{ and }1\ge\frac{\eta}{y}\ge\frac{z}{y}=\sqrt{\frac{1-\epsilon}{1+\epsilon}}\tag{4} $$ Since $f$ is convex, $f'$ is non-decreasing. Therefore, $f'(\eta)\le f'(y)\le f'(\xi)$ and so, with $(3)$ and $(4)$, we have $$ \frac{f'(y)}{y}\le\frac{f'(\xi)}{y}\le\frac{f'(\xi)}{\xi}\sqrt{\frac{1+\epsilon}{1-\epsilon}}\le2(L+\epsilon)\sqrt{\frac{1+\epsilon}{1-\epsilon}} $$ and $$ \frac{f'(y)}{y}\ge\frac{f'(\eta)}{y}\ge\frac{f'(\eta)}{\eta}\sqrt{\frac{1-\epsilon}{1+\epsilon}}\ge2(L-\epsilon)\sqrt{\frac{1-\epsilon}{1+\epsilon}} $$ Thus, for any $y>x_0\sqrt{\frac{1+\epsilon}{1-\epsilon}}$ we have $$ 2(L-\epsilon)\sqrt{\frac{1-\epsilon}{1+\epsilon}}\le\frac{f'(y)}{y}\le2(L+\epsilon)\sqrt{\frac{1+\epsilon}{1-\epsilon}} $$ Therefore, $$ \limsup\limits_{x\to+\infty}\frac{f'(x)}{x}\le2(L+\epsilon)\sqrt{\frac{1+\epsilon}{1-\epsilon}}\hspace{.25in}\text{and}\hspace{.25in}\liminf\limits_{x\to+\infty}\frac{f'(x)}{x}\ge2(L-\epsilon)\sqrt{\frac{1-\epsilon}{1+\epsilon}} $$ Since $\epsilon$ was arbitrary, we get that $$ \lim\limits_{x\to+\infty}\frac{f'(x)}{x}=2L $$

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Since leshik posted an answer and mine was a bit different, I posted mine. – robjohn Sep 1 '11 at 17:07

It is clear that once we prove, that the limit exists, the value can easily be computed using L'Hospital Rule. In fact, $\lim_{x\to\infty}\frac{f'(x)}{x}=2L.$ Let me show, that $\lim\sup\frac{f'(x)}{x}\le 2L.$ The key point is the inequality, $f(x+h)\ge f(x)+hf'(x),$ for $h\ge 0.$ This immediately follows from concavity of $f$ by differentiating with respect to $h.$ Latter is equivalent to

$$\frac{f'(x)}{x}\le \frac{(x+h)^2}{xh}\left(g(x+h)-g(x)\frac{x^2}{(x+h)^2}\right)$$ for $g(x)=\frac{f(x)}{x^2},$ $x,h>0.$ Let $\varepsilon>0$ be fixed. For all sufficiently large $x>0,$ $L-\varepsilon < g(x)<L+\varepsilon,$ thus we can estimate:

$$\frac{f'(x)}{x}\le \frac{(x+h)^2}{xh}\left(L+\varepsilon- (L-\varepsilon)\frac{x^2}{(x+h)^2}\right),$$ or $$\frac{f'(x)}{x}\le 2L+ L\frac{h}{x}+\varepsilon\frac{(x+h)^2+x^2}{xh}.$$

Now, it is easy to make $$L\frac{h}{x}+\varepsilon\frac{(x+h)^2+x^2}{xh}$$ suffitiently small by choosing $h.$ Indeed,

$$L\frac{h}{x}+\varepsilon\frac{(x+h)^2+x^2}{xh}=(L+\varepsilon)\frac{h}{x}+2\varepsilon\frac{x}{h}+2\varepsilon=2\sqrt{2\varepsilon(L+\varepsilon)}+2\varepsilon,$$ for $h=x\frac{\sqrt{2\varepsilon}}{\sqrt{L+\varepsilon}}.$ By analogy, one can prove that $\lim\inf\frac{f'(x)}{x}\ge 2L$ (just use $-h$ instead of $h$.)

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Suppose $\lim_{x \rightarrow \infty} {f'(x) \over x} = 2L$ does not hold; we will get a contradiction.

Either $\limsup_{x \rightarrow \infty} {f'(x) \over x} > 2L$ or $\liminf_{x \rightarrow \infty} {f'(x) \over x} < 2L$. We focus on the first case; afterwards we will indicate the modifications needed in the latter case. So there are $x_n \rightarrow \infty$ and $\epsilon > 0$ such that $${f'(x_n) \over x_n} > 2L + 3\epsilon$$ Since $f'(x)$ is increasing by convexity of $f$, there is a $\delta > 0$ depending only on $L$ and $\epsilon$ such that for $x \in [x_n, (1 + \delta)x_n]$ you have $${f'(x) \over x} > 2L + 2\epsilon $$ Note that $$x\bigg({f(x) \over x^2}\bigg)' = {f'(x) \over x} - 2{f(x) \over x^2}$$ Since $\lim_{x \rightarrow \infty} {f(x) \over x^2} = L$, if $n$ is large enough, then for $x \in [x_n, (1 + \delta)x_n]$ one has $$x\bigg({f(x) \over x^2}\bigg)' > \epsilon$$ Which is the same as $$\bigg({f(x) \over x^2}\bigg)' > {\epsilon \over x}$$ Integrating this from $x_n$ to $y_n = (1 + \delta)x_n$ gives $${f(y_n) \over y_n^2} - {f(x_n) \over x_n^2} > \epsilon \ln(1 + \delta)$$ Since this holds for $x_n \rightarrow \infty$, this contradicts that $\lim_{x \rightarrow \infty} {f(x) \over x^2} = L$ and we're done.

In the case where you have $\liminf_{x \rightarrow \infty} {f'(x) \over x} < 2L$, you can do the same basic argument, using $y_n = (1 - \delta)x_n$ for appropriate $\delta$ instead of $y_n = (1 + \delta)x_n$.

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