Integrable and order Let $f \in C((0,1])$ be a non-negative real-valued function, and
$$\int^1_0 f(x)dx < \infty$$
then does the following hold? $$\lim_{x \to 0} xf(x) =0$$
Any advise would be appreciated.
 A: Hints for a counterexample:


*

*if the limit $\lim_{x \to 0} x f(x)$ exists then it must be $0$ otherwise the improper integral would diverge, so the limit must not exist for a counterexample;

*$f(x)$ cannot be monotonic, otherwise the conclusion would hold per this;

*a sequence of spikes or square pulses can become arbitrarily large near $0$ and still keep the integral converging, provided they also become narrow enough.

[ EDIT ]   An explicit counterexample can be built as a "sawtooth" function with increasingly high and more narrow "teeth" towards 0. The "teeth" would need to be tall enough so that they get comparable to $\frac{1}{x}$, and narrow enough so that the improper integral still converges.

For an example, the rightmost $5$ "teeth" of such a function could be:

The above uses:
$$
\begin{align}
f(x) =
\begin{cases}
2^n \Lambda\left(2^{2n}\left(x - \frac{1}{2^n}\right)\right) & \quad x \in \left(\frac{1}{2^n} - \frac{1}{2^{2n}}, \frac{1}{2^n} + \frac{1}{2^{2n}}\right) \;\;\text{for some}\;\;n \in \mathbb{N} \text{,}\;n \ge 2  \\
0 & \quad \text{otherwise}
\end{cases}
\end{align}
$$
where $\Lambda$ is the triangular function:
$$
\begin{align}
\Lambda(x) = \max(1-|x|,0) =
\begin{cases}
1 - |x| & \quad \quad |x| \le 1 \\
0 & \quad \quad \text{otherwise}
\end{cases}
\end{align}
$$
It can be shown that the $n^{th}$ triangular "tooth":


*

*has an area of $\frac{1}{2^n}$, so the improper integral converges;

*attains a maximum of $2^n$at $x_n = \frac{1}{2^n}$, so $x_n f(x_n) = 1$ for $\forall n \ge 2$ therefore $\lim_{x \to 0} x f(x)$ cannot be $0$.
