Situation We have to continuous, differentiable functions $f(x),g(x)$ both mapping $\mathbb{R}\mapsto\mathbb{R}$. We know that
- $\exists~ \bar{x}$ such that $f(\bar{x})>g(\bar{x})$
- $0\leq f'(x)\leq g'(x)$
- $g''(x)=0\leq f''(x)$
- $\lim_{x\rightarrow\infty}[f(x)-g(x)]=0$.
In words: $f$ is larger than $g$ at a point, $g$ grows faster in $x$ than $f$, $f$ is convex while $g$ is linear and the function get arbitrarily close for $x\rightarrow\infty$.
Question Can I conclude that $f(x)>g(x)\forall x$?
Insights For $x<\bar{x}$ this is clear, as $g$ declines faster in $x$ than $f$ due to the condition of first derivatives. For $x>\bar{x}$ this should hold true. Assuming there was a $\hat{x}$ with $f(\hat{x})<g(\hat{x})$, we would need an segment of $f$ where $f''(x)\leq0$ in order to achieve the $\lim_{x\rightarrow\infty}[f(x)-g(x)]=0$. I have a problem in writing this up in a rigorous way as I have problems making the connection to the second derivative. Any help is appreciated.
Suppose there is an $x_0 > \bar x$ such that $f(x_0) < g(x_0)$, then for all $x > x_0$, $$\int_{x_0}^x f'(t)\, dt \le \int_{x_0}^x g'(t)\, dt\\f(x) - f(x_0) \le g(x) - g(x_0)\\0 < g(x_0) - f(x_0) \le g(x) - f(x)$$ So not only is $f(x) < g(x)$ for all $x > x_0$, but it never comes closer together than $g(x_0) - f(x_0)$. But we are also given that $$\lim_{x \to \infty} |f(x) - g(x)| = 0$$ But this cannot be, as $$\lim_{x \to \infty} |f(x) - g(x)| = \lim_{x \to \infty} g(x) - f(x) \ge \lim_{x \to \infty} g(x_0) - f(x_0) > 0$$
Therefore it must be that $f(x) \ge g(x)$ for all $x > \bar x$, even if $f''$ were allowed to drop below $g''$.