Show that continuous functions on $[0,1]$ satisfy this property: $\lim_{n \to \infty} n\int_0^1e^{-nx}f(x)dx$ If $f \in C[0,1]$ prove that 
$$
\lim_{n \to \infty} n\int_0^1e^{-nx}f(x)dx
$$
exists and find the limit.
I can show that $|g_n|$ are bounded by $M=\max(f)$. After some test functions I suspect that $g_n \to f(0)$ but I'm stuck!
 A: Hint. Make the change of variable $u=nx$, $du=ndx$, giving
$$
n\int_0^1e^{-nx}f(x)dx=\int_0^\infty e^{-u}f\left(\frac{u}n\right)1_{\left\{u\leq n\right\}}du
$$ observe that $\displaystyle \left|f\left(\frac{u}n\right)\right|\leq M$ over $[0,1]$ then use the dominated convergence theorem.
A: The integral (multiplied by $n$) is equal to:
$$\int_\Bbb R e^{-x} f(x/n)\chi_{[0,n]}(x)dx$$
Let $f_n(x)$ be the integrand. $(f_n)$ is a sequence of Borel functions and it's bounded uniformly by $g(x) = \|f\|_{\infty} e^{-x} \chi_{[0,\infty)}$, which is in $L^1$. Also, $f_n$ converges pointwise to $x \mapsto f(0) e^{-x} \chi_{[0,\infty)}$. Thus, by Lebesgue's dominated convergence theorem, 
$$\lim_{n \to \infty} n \int_0^1 e^{-nx} f(x) dx = f(0)\int_0^{\infty} e^{-x} dx = f(0)$$
A: A (very) slightly different approach.
$$\int_{0}^{1}n e^{-nx}f(x)\,dx = f(0)\int_{0}^{1}ne^{-nx}\,dx + \int_{0}^{1}ne^{-nx}(f(x)-f(0))\,dx \tag{1}$$
It is trivial that the first integral in the RHS converges to $f(0)$. For a given $m\geq 2$, let:
$$\delta_m = \sup_{\substack{x,y\in[0,1]\\ |x-y|\leq\frac{1}{m}}}\left|f(x)-f(y)\right|. \tag{2}$$
For any $x\in[0,1]$ we have $\left|f(x)-f(0)\right|\leq \delta_m\cdot\left\lceil\frac{x}{m}\right\rceil\leq \delta_m\left(\frac{x}{m}+1\right) $, so the second term in the RHS of $(1)$ is bounded by:
$$ \frac{\delta_m}{m}\int_{0}^{+\infty}nx e^{-nx}\,dx+\delta_m\int_{0}^{+\infty} ne^{-nx}\,dx = \frac{\delta_m}{mn}+\delta_m\leq \delta_m\cdot\left(1+\frac{1}{2n}\right) \tag{3}$$
that by the continuity of $f(x)$ can be made arbitrarily small ($\lim_{m\to +\infty}\delta_m = 0$).
