Does there exist a function $h\in C(0,1)$ such that $\lim_{x\rightarrow 0}h(x)=\infty$ and satisfying the following condintion $$ n\int_0^1 e^{-nx} h(x) dx<C,\,\forall n>>1. $$
P.s. I think such a function does exist due to the exponential decay in front. However, I am unable to construct it.
Assuming that $h$ is non-negative we have $n\int_0^{1}e^{-nx}h(x)dx=\int_0^{n}e^{-y}h(\frac y n)dy\geq \int_0^{\sqrt n}e^{-y}h(\frac y n)dy\geq c_n \int_0^{\sqrt n}e^{-y}dy$ where $c_n=\inf \{h(t): 0<t<\frac 1 {\sqrt n}\}$.Since, $c_n \to \infty$ it follows that $n\int_0^{1}e^{-nx}h(x)dx \to \infty$.