Consider three continuous functions $f,u,v:[0,1] \rightarrow (0,1)$ and a sequence of integers $0 \leq a_n \leq n $ such that \begin{equation} \lim_{n \rightarrow + \infty}{\frac{a_n}{n}}=\alpha \in (0,1) \end{equation} Suppose that there is a unique $y$ (resp. $z$) such that $u(y)=\alpha$ (resp. $v(z)=\alpha$).

I am trying to show that \begin{equation} \lim_{n \rightarrow +\infty} \frac {\int{f(x) u(x)^{a_n} (1-u (x))^{n-a_n}}dx}{\int{f(x) v(x)^{a_n} (1-v(x))^{n-a_n}} dx} = \frac{f(y)}{f(z)} \end{equation}

The reason why I suspect that this property might be valid is because an analogous result holds true in the discrete case. Take $N$ real numbers $(x_1,\cdots,x_N)$ in $[0,1]$ and three sequences $(f_1,\cdots,f_N)$, $(u_1,\cdots,u_N)$ and $(v_1,\cdots,v_N)$ with values in $(0,1)$. Suppose that there is a unique index $i \in \{1,\cdots,N\}$ (resp. $j \in \{1,\cdots,N\}$) such that $u_i=\alpha$ (resp. $v_j=\alpha$).

Then it is easy to show that for all $k \ne i$ \begin{equation*} \lim_{n \rightarrow +\infty} \dfrac{f_k u_k^{a_n} (1-u_{k})^{1-a_n}}{f_i u_{i}^{a_n} (1-u_{i})^{1-a_n}} = 0 \end{equation*}

And therefore \begin{equation*} \displaystyle \sum_{k=1}^{N}{f_k u_k^{a_n} (1-u_{k})^{1-a_n}} \displaystyle \sim f_i \alpha^{a_n} (1-\alpha)^{1-a_n} \end{equation*}

Similarly, \begin{equation*} \sum_{k=1}^{N}{f_k v_k^{a_n} (1-v_{k})^{1-a_n}} \sim f_j \alpha^{a_n} (1-\alpha)^{1-a_n} \end{equation*}

And thus \begin{equation*} \dfrac{\sum_{k=1}^{N}{f_k u_k^{a_n} (1-u_{k})^{1-a_n}}}{\sum_{k=1}^{N}{f_k v_k^{a_n} (1-v_{k})^{1-a_n}}} \sim \dfrac{f_i}{f_j} \end{equation*}

I have been unable to make any progress in the continuous case. Any advice would be much appreciated. Thanks!


1 Answer 1


The result fails. Suppose we define $a_n = n/2$ for $n$ even, $a_n = (n+1)/2$ for $n$ odd. Then we have $\alpha = 1/2.$ Simply take $f\equiv c$ for some constant $c\in (0,1).$ Finally, define

$$u(x) = \frac{1-|x-1/2|}{2},\, v(x) = \frac{1-|x-1/2|^{1/2}}{2}.$$

For both $u,v$ the unique point where these functions take on the value $\alpha = 1/2$ is $x=1/2.$

If the result were true in this case, then the limit of the ratios of the integrals would be $1.$ I'll show that this limit is actually $\infty.$

Since $f(x)$ is constant, it slides out of the integrals and cancels. Suppose $n$ is even. The expression equals

$$\frac {\int_0^1 [u(x)(1-u (x)]^{n/2} \, dx}{\int_0^1 [v(x)(1-v(x)]^{n/2} \, dx} = \frac {\int_0^1 [(1-(x-1/2)^2)/4]^{n/2} \, dx}{\int_0^1 [(1-|x-1/2|)/4]^{n/2} \, dx} $$ $$\tag 1=\frac {\int_0^1 (1-(x-1/2)^2)^{n/2} \, dx}{\int_0^1 (1-|x-1/2|)^{n/2} \, dx}$$

Now here's the thing: In the numerator of $(1)$ we have $1-(x-1/2)^2,$ which is an upside down parabola that peaks with value $1$ at $x=1/2.$ Downstairs we have $1-|x-1/2|,$ which also peaks at $x=1/2$ with value $1.$ But notice that as $x$ moves away from $1/2,$ $1-(x-1/2)^2$ decreases from $1$ at a slower rate than does $1-|x-1/2|.$ That will lead to the the numerator $\to 0$ as $n\to \infty$ at a slower rate than does the denominator.

Claim: As $n\to \infty,$ the numerator in $(1)$ is on the order of $1/\sqrt n,$ while the denominator is on the order of $1/n.$ Thus the ratio in $(1)$ is on the order of $\sqrt n \to \infty.$

Here's the proof of the claim for the denominator: By symmetry, the integral is twice the integral from $0$ to $1/2.$ For this integral, let $x=1/2-y$ to get

$$\int_0^{1/2}(1-y)^{n/2}\, dy.$$

Now let $y = z/n.$ We get

$$\frac{1}{n}\int_0^{n/2}(1-z/n)^{n/2}\, dz.$$

As $n\to \infty,$ the last integral $\to \int_0^\infty e^{-z/2}\, dz.$ Thus these integrals are on the order of $1/n$ as claimed.

I'll leave the claim for the numerators to you for the moment.


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