$f \in L^1 ((0,1))$, decreasing on $(0,1)$ implies $x f(x)\rightarrow 0$ as $x \rightarrow 0$

Question

Suppose $f\in L^1((0,1),dm)$. Suppose $f$ is also a decreasing function on $(0,1)$. Prove that

$$\lim _\limits{x \to 0} xf(x) = 0.$$

My attempt at a solution

We may assume without loss of generality that $f$ is positive, and unbounded near zero; thus we can write

$$\infty >\int_0^1 f\,dm=\sum_{n=1}^{\infty} \int_{\frac{1}{n+1}}^{\frac 1n}f\,dm.$$

Then

$$ \int_{\frac{1}{n+1}}^{\frac 1n}f\,dm \rightarrow 0 \quad \text{as}\quad n\rightarrow \infty. $$

Also since $f$ is decreasing, we have

$$f \left(\frac{1}{n}\right)\left( \frac 1n - \frac{1}{n+1}\right) \leq \int_{\frac{1}{n+1}}^{\frac 1n}f\,dm \leq f \left(\frac{1}{n+1}\right)\left( \frac 1n - \frac{1}{n+1}\right).$$

I want to say

$$ \lim _\limits{x \to 0} xf(x) = \lim _\limits{n \to \infty} \frac 1n f \left( \frac 1n \right) = 0,$$

But that's not exactly what I have in my inequalities.

Added as pointed out by @Hetebrij below,

Note that by showing $\lim _\limits{n \to \infty} \frac1n f\left( \frac 1n \right) =0$ you have not shown $\lim_\limits{x\to 0}xf(x)=0$ unless you know (1) the limit exist or (2) $xf(x)$ is decreasing or increasing.

Any hints or suggestions? Would contradiction be better suited for this problem?

UPDATE Another attempt:

Take any $n>1$. Then

$$\frac {1}{2n}f \left(\frac 1n\right) = \left(\frac 1n-\frac {1}{2n}\right)f \left(\frac 1n\right) \leq \int_{[0,1]}f(x)\chi_{\left[\frac{1}{2n},\frac 1n\right]}(x)dx.$$

Now the sequence of functions

$$f(x)\chi_{\left[\frac{1}{2n},\frac 1n\right]}(x)$$ is dominated by the integrable function $f$, and converges to $\infty\cdot \chi_{\{0\}}$. Thus by the Lebesgue Dominated Convergence Theorem, we have

$$\int_{[0,1]}f(x)\chi_{\left[\frac{1}{2n},\frac 1n\right]}(x)dx \rightarrow 0.$$

Answer 1

Let us assume first that $f\geq 0$ (which you did without much justification). Then $f (x)\geq f (x_0) $ for $x\leq x_0$ and thus $$ \int_0^{x_0} f (x)\,dx \geq \int_0^{x_0} f (x_0)\,dx = x_0 f (x_0) \geq 0. $$ But (e.g. using dominated convergence), we easily see that the left-hand side of the estimate converges to $0$. This yields the claim.

Actually, the proof above just used that for $x_0>0$ small enough, we have $f (x_0)\geq 0$.

Thus, the remaining case is that there are arbitrarily small $x_0>0$ with $f (x_0)<0$. By monotonicity, this yields $f (x)\leq 0$ for all $x\in (0,1) $. Hence, $$ 0 \geq x \cdot f (x) \geq x \cdot f (1/2) \to 0 $$ as $x \to 0$. Thus, the claim also holds in this case.