# Limit property of second derivative of bounded monotone function

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?

• Isn't $x^2 f''(x)\leq 0$ just equivalent to $f''(x)\leq 0$?
– Snaw
Commented Dec 2, 2021 at 21:32
• @Snaw It is possible to have $f'' >0$ everywhere but $\lim x^{2}f''(x) =0$. Commented Dec 2, 2021 at 23:24
• @KaviRamaMurthy If $\lim_{x\to\infty} x^2 f''(x)=0$ then surely also $\lim_{x\to\infty} f''(x)=0$? I don't see the point of the $x^2$ term
– Snaw
Commented Dec 2, 2021 at 23:43
• $\lim f''(x) \to 0$ may hold but $\lim x^{2}f''(x) \to 0$ may not hold. So OP is asking for something stronger. Commented Dec 2, 2021 at 23:51
• @KaviRamaMurthy Right, thanks.
– Snaw
Commented Dec 3, 2021 at 0:53

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.