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I'd like to prove the following statement:

Suppose $g:(0,1] \to [0,\infty)$ is a strictly decreasing, convex function satisfying $g(1) = 0$ and $\lim_{t \downarrow 0} g(t) = \infty$. Show that $$ \lim_{t \downarrow 0} \frac{g(t)}{g'(t)} = 0 $$

My idea is to rewrite the limit as $$ \lim_{t \downarrow 0} \left(\frac{d\log g(t)}{dt}\right)^{-1} $$ and show \begin{align}\tag{1} \lim_{t \downarrow 0} \frac{d\log g(t)}{dt} = \infty \end{align} I thought I could do this by proving the statement $$ \lim_{x \to x_0} f(x) = \infty, \; f'(x) > 0 \quad \Rightarrow \quad \lim_{x \to x_0} f'(x) = \infty $$ But upon more thought, I realize this statement is not true unless we additionally assume $f$ is convex. Since we did not assume $\log g$ is convex, I'm not sure if this idea will still work.

Any ideas?

Thanks!

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Because $g$ is strictly decreasing, we know that $g' \le 0$, and doesn't equal $0$ on any interval. Because $g$ is convex, we know that $g'$ is increasing.

Suppose $\frac {g'(t)}{g(t)}$ does not diverge to $-\infty$ as $t \to 0^+$. Then there must be an $M$ such that for any $t > 0$, there is an $a$ with $0 < a < t$ such that $g'(a)/g(a) > M$. (This is the negation of the definition of $\lim_{t\downarrow0}g'(t)/g(t) = -\infty$.)

Since $g'$ is increasing and $g$ is decreasing, $g'(t) \ge g'(a) > Mg(a) > Mg(t)$. Therefore for all $t$, $$\frac{g'(t)}{g(t)} > M$$ Choose some arbitrary fixed point $b \in (0,1)$, and integrate: $$\int_x^b \frac{g'(t)}{g(t)}dt > M\int_x^b dt$$ $$\log g(b) - \log g(x) > M(b - x)$$ $$\log g(b) - M(b - x) > \log g(x)$$ $$g(b)e^{M(x - b)} > g(x)$$

But the LHS is bounded for $x \in (0,1]$, contradicting that $g(0^+) = \infty$. Hence $\frac {g'(t)}{g(t)} \to -\infty$ as $t\to 0^+$

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$\lim_{t\downarrow 0} (\ln g(t))'=0$.

Perhaps you can integrate it, and you get

$\lim_{t\downarrow 0} \ln g(t) = \text{const}$

Since $g(t)=\infty$ when $t\to 0$ and lim and ln can switch places since ln is continuous function, you really get that $\text{const} = \infty$.

Was that helpful?

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    $\begingroup$ Here is a guide on how to use mathjax. You will find you answers are better received when they are easy to read. [meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Paul Sinclair Feb 20 '16 at 0:14
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    $\begingroup$ Also, your work is going the wrong way and to the wrong limits. He knows that $\lim_{t\downarrow 0} \ln g(t) = \infty$. What he needs to show is that $$\lim_{t\downarrow 0} \left|\frac{d(\log g(t))}{dt}\right| = \infty$$ $\endgroup$ – Paul Sinclair Feb 20 '16 at 0:24

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