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How can prove this question?

Let $f: [0.1]\to\mathbb{R}$ be a convex function of class $C^1$, with $f'(0)> 0$.

Prove that

$$\lim_{x \to +\infty}f(x)=+\infty$$

And that the limit exists finite

$$\lim_{x \to +\infty}\frac{x}{f(x)}$$

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  • $\begingroup$ $f$ should be defined on $\mathbb R$ or at least $\mathbb R_+$, since you talk about a limit at $+\infty$. $\endgroup$ – Davide Giraudo Jan 20 '12 at 9:40
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    $\begingroup$ As always, no source, no motivation, no mention of failed approaches. $\endgroup$ – Did Jan 20 '12 at 9:47
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A standard result is that a $C^1$ convex function on $\mathbb R$ (or any open interval) has an increasing derivative. So $$f(x)=\int_0^xf'(t)dt+f(0)\geq\int_0^xf'(0)dt+f(0)=xf'(0)+f(0),$$ and we get the result since $f'(0)>0$.

For the second question, put $g(x)=\frac{f(x)}x$ for $x>0$. Then $$g'(x)=\frac{f'(x)x-f(x)}{x^2}.$$ Now fix $x_0>0$ such that $f(x_0)>0$. Then for $x\geq x_0$ $$g'(x)=\frac{f'(x)(x-x_0)+f'(x)x_0-f(x)}{x^2}=\frac{f'(x)(x-x_0)+f'(x)x_0-\int_{x_0}^xf'(t)dt+f(x_0)}{x^2},$$ and since $f'(x)(x-x_0)-\int_{x_0}^xf'(t)dt\geq 0$ and $f'(x)x_0+f(x_0)\geq 0$, we get that $h$ is increasing on $[x_0,+\infty[$ so $x\mapsto \frac x{f(x)}$ is decreasing on $[x_0,+\infty[$. Therefore, the limit $\lim_{x\to +\infty}\frac x{f(x)}$ exists and is finite since $\frac x{f(x)}> 0$ for $x$ large enough.

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