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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.

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.

-Mario-

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    Hint: the Mean Value Theorem says that for some $\xi$ so that y<\xi,$\frac{f(x)-f(y)}{x^2-y^2}=\frac{f'(\xi)}{2\xi}$ If that is not enough, I have a complete proof I can post.2011-09-01

3 Answers 3

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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 leshi$k$ posted an answer and mine was a bit different, I posted mine.2011-09-01
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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) 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 $y_n = (1 + \delta)x_n$.