Why does $\int_0^T ne^{n(x-T)}f(x)$ converge to $f(T)$? Let $f(x)$ be a nice function. Consider the integral $$\int_0^T ne^{n(x-T)}f(x)\ dx.$$
Apparently, this converges to $f(T)$ as $n\to\infty$. I can't see why. I've tried doing integration by parts, but it doesn't lead anywhere. I've put this in Wolfram Alpha for various choices of $f(x)$ and it seems to be true.
I can see that the factor $ne^{n(x-T)}$ seems to concentrate its mass around $x=T$ as $n\to\infty$, but then this would seem to suggest that the integral is almost $n\int_0^T \chi([2-\epsilon,2]) f(x)$. Not really sure what's going on!
Why does this integral converge to $f(T)$?
EDIT: I tried integration by parts as follows. Let $v=f(x)$ and $du=ne^{-n(x-T)}$. Then
$$\int_0^T ne^{n(x-T)}f(x)\ dx = e^{n(x-T)}f(x)\Bigg|_0^T - \int_0^T e^{n(x-T)}f'(x)$$
$$=f(T) - e^{-nT}f(0)- \int_0^T e^{n(x-T)}f'(x),$$
which doesn't seem to lead anywhere.
 A: Note that
$$\int_0^T ne^{n(x-T)}f(x)\ dx = \int_0^T ne^{-nx}f(T-x)\ dx. $$
Assuming only that $f$ is continuous, we can show this converges to $f(T)$ using calculus. No assumption about differentiability is required.
We have
$$I_n = \int_0^T ne^{-nx}f(T- x)\, dx= \int_0^T ne^{-nx}[f(T- x) - f(T)] \, dx + f(T)\int_0^T ne^{-nx} \, dx \\ = \int_0^c ne^{-nx}[f(T-x) - f(T)] \, dx +  \int_c^T ne^{-nx}[f(T-x) - f(T)] \, dx + f(T)(1- e^{-nT}),$$
and
$$\left| I_n - f(T)\right| \leqslant \int_0^c ne^{-nx}|f(T-x) - f(T)| \, dx +  \int_c^T ne^{-nx}|f(T-x) - f(T)| \, dx + |f(T)|e^{-nT}$$
For any $\epsilon > 0,$ choose $c$ sufficiently small such that $|f(T-x) - f(T)| < \epsilon$ for $ 0 < x < c$.
Then
$$|I_n - f(0)| \leqslant \epsilon(1 - e^{-nc}) + (2\sup_{x \in [0,T]} |f(x)|)(e^{-nc}- e^{-nT}) + |f(T)|e^{-nT} .$$
Taking the limit as $n \to \infty$ we get
$$\lim_{n \to \infty}|I_n - f(0)| \leqslant \epsilon.$$
Since $\epsilon$ can be  arbitrarily small, it follows that $\lim I_n = f(T).$
Another approach would be to use the dominated convergence theorem.
A: this is an approach rather than an answer:
First do the case when $f$ is constant. This should be easy.
Now subtract $f(T)$ to reduce to the case where $f(T)=0.$ 
Pick any $y < T,$ show that the integral up to $y$ goes to zero. This should follow from the dominated convergence theorem since the integrands are uniformly bounded on such an interval and go to zero point wise. 
At $T$ you can write $f(x) = (x-T) g,$ with $g$ bounded by Taylor's theorem. Use this to bound the integrands near 1 and again get it to go zero.
