Limit Brownian Bridge Integral

Question

As a solution of the Brownian Bridge SDE, we arrive at the solution \begin{align} X_t = (1-t) \int_0^t \frac{1}{1-s}\ dB_S \end{align} defined for $0 \leq t <1$. In order to show that for any $g \in C[0,1]$ \begin{align} \lim_{t \uparrow 1}\ (1-t) \int_0^t \frac{g(s)}{(1-s)^2}\ ds = g(1), \end{align} I am considering two cases.

Case 1:
\begin{align} \lim_{t \uparrow 1}\ \int_0^t \frac{g(s)}{(1-s)^2}\ ds = \lim_{t \uparrow 1}\ \frac{1}{1-t} = \infty, \end{align} such that we apply l'Hôpital's rule and find that \begin{align} \lim_{t \uparrow 1}\ \frac{\frac{g(t)}{(1-t)^2}}{\frac{1}{(1-t)^2}} = g(1). \end{align}

Case 2: \begin{align} \lim_{t \uparrow 1}\ \int_0^t \frac{g(s)}{(1-s)^2}\ ds \neq \lim_{t \uparrow 1}\ \frac{1}{1-t} = \infty. \end{align} Now, clearly \begin{align} (*) = \lim_{t \uparrow 1}\ \frac{ \int_0^t \frac{g(s)}{(1-s)^2}\ ds }{\frac{1}{1-t}} \to 0 \qquad \text{since} \qquad \lim_{t \uparrow 1}\ \frac{1}{1-t} = \infty. \end{align} However, how to show that $(*) \to g(1)$ as well?

Answer 1

NOTE:

Referring to the note at the end of the General Proof Section of THIS ARTICLE, L'Hospital's Rule states that given two functions $f$ and $g$ that are differentiable in an open neighborhood of $\xi$ with $g'(x)\ne 0$ in that neighborhood, such that $\lim_{x\to \xi}|g(x)|=\infty$, then if the limit $\lim_{x\to \xi}\frac{f'(x)}{g'(x)}$ exists, whether finite or infinite, then we have

$$\lim_{x\to \xi}\frac{f(x)}{g(x)}=\lim_{x\to \xi}\frac{f'(x)}{g'(x)}$$

Note carefully that nowhere in the statement of L'Hospital's Rule is the assumption that $\lim_{x\to \xi}f(x)$ even exists.

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Now, I thought it might be instructive to present an approach that does not use L'Hospital's Rule. Rather, we use the fact that $g(s)$ is continuous on $[0,1]$, and therefore uniformly continuous there.

As a first step, we write

$$\begin{align} (1-t)\int_0^t \frac{g(s)}{(1-s)^2}\,ds&=(1-t)\left(\int_0^t \frac{g(s)-g(t)}{(1-s)^2}\,ds+g(t)\int_0^t \frac{1}{(1-s)^2}\,ds\right)\\\\ &=t\,g(t)+(1-t)\int_0^t \frac{g(s)-g(t)}{(1-s)^2}\,ds\tag 1 \end{align}$$

Second, from the uniform continuity of $g$, given any $\epsilon>0$, there exists a number $\delta>0$ such that $|g(s)-g(t)|<\epsilon$ whenever $t-\delta <s<t$. For that $\delta$, we now take $1> t>1-\delta$. Then, we have

$$\begin{align} \left|(1-t)\int_0^t \frac{g(s)-g(t)}{(1-s)^2}\,ds\right|&=\left|(1-t)\int_0^{t-\delta} \frac{g(s)-g(t)}{(1-s)^2}\,ds+(1-t)\int_{t-\delta}^t \frac{g(s)-g(t)}{(1-s)^2}\,ds\right|\\\\ &\le (1-t)\int_0^{t-\delta} \frac{|g(s)-g(t)|}{(1-s)^2}\,ds+(1-t) \int_{t-\delta}^t \frac{|g(s)-g(t)|}{(1-s)^2}\,ds\\\\ &\le \frac{2||g||_{\infty}(1-t)(1-\delta)}{\delta}+\epsilon\left(1-\frac{(1-t)}{\delta}\right) \tag 2 \end{align}$$

Third, letting $t\to 1^-$ in $(2)$ we see that for any $\epsilon>0$, $\lim_{t\to 1^-}\left|(1-t)\int_0^t \frac{g(s)-g(1)}{(1-s)^2}\,ds\right|<\epsilon$, and hence the limit is zero.

Using this result in $(1)$ results in the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{t\to 1^-}(1-t)\int_0^t \frac{g(s)}{(1-s)^2}\,ds=g(1)}$$

And we are done!