User:Msalins/sandbox

Formal statement and proof:

Assume that $$ f(x) $$ is a twice differentiable function on $$ [a,b] $$ with $$ x_0 \in [a,b] $$ the unique point such that $$ f(x_0) = \max_{[a,b]} f(x) $$. Assume additionally that $$ f''(x_0)<0 $$ and that for any $$ \delta >0 $$, $$ \sup_{[a,b] \setminus (x_0 - \delta, x_0 + \delta)} f(x) < f(x_0) $$.

Then,

$$ \lim_{n \to +\infty} \left( \frac{\int_a^b e^{nf(x)} dx}{\left( e^{nf(x_0)}\sqrt{\frac{2 \pi}{n (-f''(x_0))}} \right)} \right) =1 $$

Proof:

Lower bound:

Let $$ \epsilon > 0 $$. Then by the continuity of $$ f $$ there exists $$ \delta >0 $$ such that if $$ |x_0-c|< \delta $$ then $$ f(c) \ge f''(x_0) - \epsilon. $$. By Taylor's Theorem,  for any $$ x \in (x_0 - \delta, x_0 + \delta) $$, $$f(x) \ge f(x_0) + \frac{1}{2}(f''(x_0) - \epsilon)(x-x_0)^2  $$.

Then we have the following lower bound:

$$ \int_a^b e^{n f(x) } dx \ge \int_{x_0 - \delta}^{x_0 + \delta} e^{n f(x)} dx \ge e^{n f(x_0)} \int_{x_0 - \delta}^{x_0 + \delta} e^{\frac{1}{2} (f(x_0) - \epsilon)(x-x_0)^2} dx \ge e^{n f(x_0)} \sqrt{\frac{1}{n (-f(x_0) + \epsilon)}} \int_{-\delta \sqrt{n (-f(x_0) + \epsilon)} }^{\delta \sqrt{n (-f(x_0) + \epsilon)} } e^{-\frac{1}{2}y^2} dy $$

where the last inequality was obtained by a change of variables $$ y= \sqrt{n (-f''(x_0) + \epsilon) (x-x_0)}$$. Remember that $$ f''(x_0)<0 $$ so that is why we can take the square root of its negation.

If we divide both sides of the above inequality by $$ e^{nf(x_0)}\sqrt{\frac{2 \pi}{n (-f''(x_0))}} $$ and take the limit we get:

$$ \lim_{n \to +\infty} \left( \frac{\int_a^b e^{nf(x)} dx}{\left( e^{nf(x_0)}\sqrt{\frac{2 \pi}{n (-f''(x_0))}} \right)} \right) \ge \lim_{n \to +\infty} \frac{1}{\sqrt{2 \pi}} \int_{-\delta\sqrt{n (-f(x_0) + \epsilon)} }^{\delta \sqrt{n (-f(x_0) + \epsilon)} } e^{-\frac{1}{2}y^2} dy \sqrt{\frac{-f(x_0)}{-f(x_0) + \epsilon}} = \sqrt{\frac{-f(x_0)}{-f(x_0) + \epsilon}} $$

since this is true for arbitrary $$ \epsilon $$ we get the lower bound:

$$ \lim_{n \to +\infty} \left( \frac{\int_a^b e^{nf(x)} dx}{\left( e^{nf(x_0)}\sqrt{\frac{2 \pi}{n (-f''(x_0))}} \right)} \right) \ge 1 $$

Upper bound:

The proof of the upper bound is similar to the proof of the lower bound but there are a few annoyances. Again we start by picking an $$ \epsilon >0 $$ but in order for the the proof to work we need $$ \epsilon $$ small enough so that $$ f''(x_0) + \epsilon < 0 $$. Then, as above, by continuity of $$ f $$ and Taylor's Theorem we can find $$ \delta>0 $$ so that if $$ |x-x_0| < \delta $$, then $$ f(x) \le f(x_0) + \frac{1}{2} (f(x_0) + \epsilon)(x-x_0)^2 $$. Lastly, by our assumptions there exists an $$ \eta >0 $$ such that if $$ |x-x_0|\ge \delta $$, then $$ f(x) \le f(x_0) - \eta $$.

Then we can calculate the following upper bound:

$$ \int_a^b e^{n f(x) } dx \le \int_a^{x_0-\delta} e^{n f(x) } dx + \int_{x_0-\delta}^{x_0 + \delta} e^{n f(x) } dx + \int_{x_0 + \delta}^b e^{n f(x) } dx \le (a-b)e^{n (f(x_0) - \eta)} + \int_{x_0-\delta}^{x_0 + \delta} e^{n f(x) } dx $$

$$ \le (a-b)e^{n (f(x_0) - \eta)} + e^{n f(x_0)} \int_{x_0-\delta}^{x_0 + \delta} e^{\frac{1}{2} (f(x_0)+\epsilon)(x-x_0)^2} dx \le (a-b)e^{n (f(x_0) - \eta)} + e^{n f(x_0)} \int_{-\infty}^{+\infty} e^{\frac{1}{2} (f(x_0)+\epsilon)(x-x_0)^2} dx $$

$$ \le (a-b)e^{n (f(x_0) - \eta)} + e^{n f(x_0)} \sqrt{\frac{2 \pi}{n (-f''(x_0) - \epsilon)}} $$

If we divide both sides of the above inequality by $$ e^{nf(x_0)}\sqrt{\frac{2 \pi}{n (-f''(x_0))}} $$ and take the limit we get:

$$ \lim_{n \to +\infty} \left( \frac{\int_a^b e^{nf(x)} dx}{\left( e^{nf(x_0)}\sqrt{\frac{2 \pi}{n (-f''(x_0))}} \right)} \right) \le \lim_{n \to +\infty} \left( (a-b) e^{-\eta n} \sqrt{\frac{n (-f(x_0))}{2 \pi}} + \sqrt{\frac{-f(x_0)}{-f''(x_0) - \epsilon}} \right) = \sqrt{\frac{-f(x_0)}{-f(x_0) - \epsilon}} $$

Since $$ \epsilon $$ is arbitrary we get the upper bound:

$$ \lim_{n \to +\infty} \left( \frac{\int_a^b e^{nf(x)} dx}{\left( e^{nf(x_0)}\sqrt{\frac{2 \pi}{n (-f''(x_0))}} \right)} \right) \le 1 $$

And combining this with the lower bound gives the result.