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If $f$ is strictly increasing and strictly convex (or $f'>0$ & $f''>0$), then $$\lim_{x\rightarrow∞}{f(x)}=∞$$

Is this statement true?

If this statement is true, how can I prove?

The Phenotype
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student lee
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    You might use the fact that convex functions lie above their tangents (i.e, for any fixed point $a$, $f(x) \geq f(a) + f'(a)(x-a)$). – Clement C. May 14 '13 at 02:36

2 Answers2

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Let $m=f'(0)$ and set $g(x)=mx+f(0)$. We have $f(x)\ge g(x)$ for all $x>0$ and $\lim_{x\rightarrow \infty}g(x)=\infty$.

vadim123
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vadim123's answer is clinical, I like it. A more traditional proof would be:

For $\forall x_0$, consider for $a>0$, $f(x_0 + a)$'s Taylor expansion at $x_0$: $$ f(x_0+a) = f(x_0) + af'(\xi_0) $$ for some $\xi_0 \in (x_0,x_0+a)$. Since $f''>0$ implies $f'$ is strictly increasing, we have $f'(\xi_0) > f'(x_0)= b >0$, this gives us: $$ f(x_0+a) > f(x_0) + ab. $$ Now consider $f(x_0 + 2a)$'s Taylor expansion at $(x_0+a)$, similar argument as above gives: $$ f(x_0+2a) = f(x_0+a) + af'(\xi_1)> f(x_0) + ab + af'(\xi_1) > f(x_0) + 2ab $$ Deduction yields: $$ f(x_0 + na)> f(x_0) + nab. $$ Back to $f$, for $a>0$ fixed: $$ \lim_{x\to \infty} f(x)= \lim_{n\to \infty} f(x_0+ na) \geq \lim_{n\to \infty} \big( f(x_0) + nab\big) = \infty. $$

Shuhao Cao
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