Assume on the contrary, $f'(x)\geq 0$, and since $f''(x)\leq 0$, then $f(x)$ is concave down and decreasing, so $f(x)$ will eventually less than $0$, contradict to $f(x)>0$, but how can I prove that $f(x)$ will eventually less than $0$ formally? thanks!
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Suppose $f'(a) <0$. Since $f'$ is non-increasing we get $f'(x) \leq f'(a)$ for all $x \geq a$. Hence $0 <f(x)=f(a)+\int_a^{x} f'(t)dt \leq f(a)+f'(a)(x-a)$ for $x >a$. As $x \to \infty$ RHS tends to $-\infty$ leading to a contradiction.
Kavi Rama Murthy
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I am bit confused as if we take $f(x) = - (x-1)(x-2) $ when $x\in (1,2)$, then $f(x) $ doesn't satisfy the above in $(1,2)$. $f'(x) $ changes sign at $x=3/2$ – Koro Apr 25 '20 at 23:48
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I don't understand. First of all you have to take the domain as $(0,\infty)$; the fact that $f''(x) \leq 0$ on this entire domain is important for the proof. – Kavi Rama Murthy Apr 25 '20 at 23:51
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– forgottenarrow Apr 25 '20 at 22:24For sufficiently large $x$, $g(x) < 0$ (Hint: try explicitly writing $g(x)$ in terms of $f(a)$ when $x > a$).
$f(x) \leq g(x)$ so that $f(x) < 0$ when $x$ is sufficiently large.