Show that an integral can be made as small as possible.

0 votes
asked Apr 6, 2014 by Alex Strife

Consider a function $mu(s)$ satisfying the following properties:

  1. $mu(s) in C^0((0,+infty))$,
  2. $mu(s) > 0$ and $mu(s)$ is increasing in $s in (0,+infty)$,
  3. $displaystyle int_0^1 {dfrac{mu(s)}{s} ds} < +infty.$

Show that for every $epsilon > 0$, there is a $delta >0 $ such that$$displaystyle int_0^delta {dfrac{mu(s)}{s} ds} < epsilon.$$

My idea: Set $displaystyle F(delta) := int_0^delta {dfrac{mu(s)}{s} ds}$. Then the conclusion follows if we can show that
$$ lim_{delta to 0^+} {F(delta)} = 0 $$This, in turn, will follow if $F$ is continuous on $[0,delta)$. There will be no problem if $mu(s) /s$ were bounded on $[0,delta)$ but this is not always the case; take $mu(s) = s^a (0

1 Answer

0 votes
answered Apr 6, 2014 by Ant

Well

$$int_{0}^delta frac{mu(s)}{s} + int_{delta}^1 frac{mu(s)}{s} = int_{0}^1 frac{mu(s)}{s} = l in mathbb{R} (star)$$

Now since $f(delta) = int_delta^1 frac{mu(s)}{s}$ is continuos, and since $f(0) = l$, $f(1) = 0$, it implies that $f(delta)$ will eventually take any value between $0$ and $l$, and in particular $$forall epsilon > 0 exists delta : f(delta) = l - epsilon$$

And so substituting in $(star)$

$$int_0^delta frac{mu(s)}{s} = l - f(delta) = epsilon$$

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