Show $[\log(V(r))]' \to -\infty$ for $V(r) > 0$ monotone decreasing and $r$ tending to boundary of support of $V$

Question

Suppose $V \in C^1([0, \infty);[0,\infty))$, not identically zero, and is compactly supported and has $V' < 0$ on the interior of the support of $V$.

I would like to show that $\frac{V'(r)}{V(r)}$ tends to minus infinity as $r$ tends to $a$, the boundary of the support of $V$.

I'm not completely sure that the statement is true. I haven't made much progress so far. There aren't many assumptions so certainly if there is a proof it can't be too complicated.

We have

$$\frac{V'(r)}{V(r)} = \frac{V'(r)}{-\int^a_rV'(s)ds}.$$

So somehow we need to show $\int^r_aV'(s)ds$ vanishes faster than $V'(r)$ as $r \to a$. If we impose the additional assumption that $V'$ is monotone increasing, then we will have $-\int^a_rV'(s)ds \le -V'(r)(r-a)$, which gives the result we want.

Another perspective that could be helpful is that $V'(r)/V(r) = [\log(V(r))]'$ and clearly $\log(V(r)) \to -\infty$ as $r \to a$.

Answer 1

By considering $f(x) = V(a-x)$ the question can be formulated as follows:

Let $f: [0, a] \to \Bbb R$ be continuously differentiable with $f(0) =f'(0) = 0$ and $f'(x) > 0$ for $0 < x \le a$. Does it follow that $ \lim_{x \to 0} \frac{f'(x)}{f(x)} = +\infty $?

Note that $f'/f$ can not be bounded above in any neighborhood of zero, otherwise it would be identically zero, see for example A non-trivial, non-negative, function bounded below by its derivative with $f(0)=0$?.

But the answer to the above question is NO. The following example shows that $f'/f$ can “oscillate” and can take arbitrarily small values near the origin:

Let $g:(0, a] \to \Bbb R$ be a differentiable function with the following properties:

• $g(x) \ge 1/x$,
• $g'(x) < 0$,
• $g'(1/n) >- 1/n$ for $n=1, 2, 3, \ldots$.

Such a function can for example be constructed by modifying $g(x) = 1 + 1/x$ in small intervals around $1/n$ to satisfy the last condition.

Now define $f$ as $f(0) = 0$ and $f(x) = e^{-g(x)}$ for $0 < x \le a$.

From $0 < f(x) \le e^{-1/x}$ for $0 < x \le a$ one can conclude that $f$ is continuous and differentiable at $x=0$, with $f'(0) = 0$. Also $f'(x) = -g'(x)e^{-g(x)}> 0$ for $0 < x \le a$. But $$ \frac{f'(x)}{f(x)} = -g'(x) $$ does not converge to $+\infty$ for $x \to 0$.