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Suppose that a function $f: \left( a, \infty \right) \to \mathbb{R}$, where $a \in \mathbb{R}$, is twice differentiable and convex and that $f$ has an asymptote at $+ \infty $. I want to prove that the graph of $f$ never goes below the asymptote. The problem seems to be trivial but I can't solve it. Does anyone have any ideas?

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    Stricly above, or $\ge?$ – 2017-01-19
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    In case of a linear function the graph is actually the asymptote, so $\ge $. – 2017-01-19

2 Answers 2

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Suppose the asymptote is $y = c x + d$. This says that the function $g(x) = f(x) - c x - d \to 0$ as $x \to +\infty$. Note that $g(x)$ is also convex.

Suppose $g(t) < 0$ for some $t$. Since $g(x) \to 0$ as $x \to +\infty$, there is some $s > t$ such that $g(s) > g(t)/2$. By convexity, for $x > s$ we have

$$ g(x) \ge g(s) + (x-s) \frac{g(s)-g(t)}{s-t} > \frac{g(t)}{2} - \frac{(x-s)g(t)}{2(s-t)} $$ and the right side goes to $+\infty$ as $x \to +\infty$, contradicting the assumption that $g(x) \to 0$ as $x \to +\infty$.

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Let $y=mx+b$ be the asymptote. Suppose that $f(x_0)L$, we have $|f(x)-(mx+b)|<\epsilon$. Thus for $x>\max\{L,x_0\}$, we have that $$mx+b-f(x)\le |f(x)-(mx+b)|<|f(x_0)-(mx_0+b)|=mx_0+b-f(x_0)$$ and so $$\frac{f(x)-f(x_0)}{x-x_0}>m $$ It follows that $f'(\xi)>m$ for some $\xi\in(x_0,x)$. As $f'$ is non-decreasing, it follows that $f(x)\ge f(\xi)+(x-\xi)f'(\xi)$ for all $x>\xi$. As $f'(\xi)>m$, we can find $x_1$ such that $f(x)>mx+b+1$ follows for all $x>x_1$, contradicting the asymptote property.