Existence of $f\in L^2[0,1]$ such that $\lim_{n\to\infty}\sqrt{n}\int_{0}^{\frac{1}{n}}f(x)dx=k\ne 0$ exist. Does there exists a square integrable function $f$  i.e. $f\in L^2[0,1]$ on $[0,1]$ such that $\lim_{n\to\infty}\sqrt{n}\int_{0}^{\frac{1}{n}}f(x)dx$ exist and non-zero.
Giving any idea will be appreciated.
(Previously, I was asking a uniformly continuous function f which is trivial but I need that updated one.)
 A: No.
If there is a constant $M$ with $|f| \le M$ then the answer is no because
$$\left| \sqrt{n} \int_0^{1/n} f(x) \, dx \right| \le \frac{M}{\sqrt n}.$$
If on the other hand $f$ is unbounded then for each $M \in \mathbf N$ you can define the truncation
$$ f_M(x) = \left\{ \begin{array}{cl} f(x) & |f(x)| \le M \\ 0 & |f(x)| > M.\end{array} \right.$$
By the earlier remark you have $\displaystyle \sqrt n \int_0^{1/n} f_M(x) \, dx \to 0$ for any $M$. On the other hand, by Holders inequality you have
\begin{align*}\left| \int_0^{1/n} f_M(x) \, dx - \int_0^{1/n} f(x) \, dx\right| &\le \int_0^{1/n} |f_M(x) - f(x)| \, dx \\
&\le \left( \int_0^{1/n} \, dx \right)^{1/2} \left( \int_0^{1/n} |f_M(x) - f(x)|^2 \, dx \right)^{1/2} \\ &\le \frac{1}{\sqrt{n}} \|f_M - f\|_2 
\end{align*}
so that
$$\left| \sqrt n \int_0^{1/n} f_M(x) \, dx - \sqrt n\int_0^{1/n} f(x) \, dx\right| \le \|f_M - f\|_2.$$
Now, by (for instance) LDCT you have $\|f_M - f\|_2 \to 0$ as $M \to \infty$, so that $\|f_M - f\|$ can be made arbitrarily small. This forces
$$\sqrt n\int_0^{1/n} f(x) \, dx \to 0.$$
