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I got this problem:

Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function that satisfy $\forall x\in\mathbb{R},0\leq f'(x)$ and $0\leq f''(x)$

Prove that if $\exists a\in\mathbb{R}$ such that $0<f'(a)$, Then $lim_{x\to\infty}f(x)=\infty$

I tried to show that for each $0<M$ there exist $0<N$ such that for all $N<x$, $M<f(x)$, but I got stuck choosing an $N$ that satisfies the condition.

Thanks on any hint.

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Because $f''\geq 0$, you know that $f'(x)\geq f'(a)>0$ for $x\geq a$. Then, for $x\geq a$, $$ f(x)=f(a)+\int_a^xf'(s)ds\geq f(a)+\int_a^xf'(a)ds=f(a)+f'(a)(x-a)\to\infty $$ as $x\to\infty$.


As for $M$ and $N$, from the above, you can just choose $N>a+\frac{M-f(a)}{f'(a)}$.

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The constraints on the first and second derivative imply that $f$ is convex and increasing, hence it is necessarily unbounded since: $$\forall x,x_0,\quad f(x)\geq f(x_0)+f'(x_0)(x-x_0).$$

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Hints Notice the following-

  • there exists a such that $f'(a)>0$
  • $f''(x)> 0 \space \forall \space x \implies f'(x)$ is increasing
  • $f'(x) > f'(a) \space \forall \space x\geq a$
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