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Suppose $f:\mathbb R \rightarrow \mathbb R$ is twice continuously differentiable, bounded and monotone. Is it possible to show that $\lim\inf_{x\rightarrow \infty} x^{2}f''(x)\leq 0$?
This is established in the following question when $\lim_{x\rightarrow \infty} xf'(x)= 0$:
Limit property of second derivative of bounded function.
The proof offered here relies on this property of the first derivative. I am wondering if it is possible to show without this property?
Suppose that $f$ is twice continuously differentiable, monotone and bounded on $\mathbb R$ (although it's enough for $f$ to have this property on $[a,\infty)$ for any $a$).
Suppose, by way of contradiction, that $\liminf_{x \to \infty} x^2f''(x)>0$. This implies that there exist $R,\epsilon>0$ such that for $r>R$, $r^2f''(r)>\epsilon$. This is because if this were not the case, then inductively one can find a sequence of points $r_n$ such that $r_n^2f''(r_n)< \frac 1n$ for each $n$ with $r_n \to \infty$, and therefore $\liminf x^2f''(x) \leq 0$, a contradiction.
Now, if $R,\epsilon$ exist as above, then in particular, $f''(r)>0$ for $r>R$. Since $f'' = (f')'$ , the positivity of $f''$ implies that $f'$ is monotone increasing on $[R,\infty)$.
Fix some $T>R$.
Now, note that $f$ is bounded and monotone. Therefore, we know that $\lim_{x \to \infty} f(x) = L$ exists and is finite. From the FTC , we get $L-f(T) = \int_T^{\infty} f'(t)dt$. Now, as $f'$ is monotone, $\lim_{x \to \infty} f'(x)$ exists as a possibly infinite number : however, note that $\int_T^{\infty} f'(t)dt$ also exists. Combining these two facts, it must happen that $\lim_{x \to \infty} f'(x) = 0$.
Now, as $f'$ is monotone increasing, it follows that $f'(x) \leq 0$ for all $x \in [T,\infty)$. Therefore, $f$ is monotone decreasing on $[T,\infty)$, if the contradiction holds.
We've actually not used the complete strength of the hypotheses present to us. We will use it, soon enough. I now present a part of an argument that I saw in the lead-up to the Karamata monotone density theorem.
Fix $b>1$ and let $x$ vary over large enough values such that $x,bx \in [T,\infty)$. Now, we use the FTC to write $$ \int_{x}^{bx} f'(t)dt = f(bx) - f(x) $$
By the monotonic increasing nature of $f'$, we get that $$ (b-1)x f'(bx) = \int_{x}^{bx} f'(bx)dt \geq \int_{x}^{bx} f'(t)dt \geq \int_{x}^{bx} f'(x)dt = (b-1)xf'(x) $$
for all $x$, from where we get $$ xf'(bx) \geq \frac{f(bx) - f(x)}{b-1} \geq xf'(x) $$
for all $x$. Taking the $\limsup$ as $x \to \infty$ for the second inequality yields $$ \limsup_{x\to\infty} xf'(x) \leq \limsup_{x \to \infty} \frac{f(bx)-f(x)}{b-1} = \frac 1{b-1}\left[\lim_{x \to \infty} f(bx) - \lim_{x \to \infty} f(x)\right] = 0 $$
Take the first inequality and rewrite it as $$ (bx)f'(bx) \geq \frac{b(f(bx) - f(x))}{b-1} $$
Now, we get by taking the $\liminf$ as $x \to \infty$ on both sides : $$ \liminf_{x \to \infty} xf'(x) = \liminf_{x \to \infty} bxf'(bx) \geq \liminf_{x \to \infty} \frac{b(f(bx) - f(x))}{b-1} \geq 0 $$
(The first equality is true by using the very simple $y = \frac xb$ or $y=xb$ substitution for any sequence leading to the smallest limit point). Therefore, the hypothesis imply that $\liminf_{x \to \infty} xf'(x) \geq 0 \geq \limsup_{x \to \infty} xf'(x)$ i.e. that $\lim_{x \to \infty} xf'(x) = 0$.
From here, one can finish using the linked MSE post, which proves that $\lim_{x \to \infty} xf'(x)=0$ implies a contradiction to the statement $\liminf_{x \to \infty} x^2f''(x)>0$. Hence, we are done.
