Is it true that if $g$ is monotone on $(0,1]$ and $\int_0^1 x^k g$ exists for some $k>-1$, then $\displaystyle\lim_{x\to0} x^{k+1} g(x)=0$?

Question

Is it true that if $g$ is monotone on $(0,1]$ and $\int_0^1 x^k g$ exists for some $k>-1$, then $\lim_{x\to0} x^{k+1} g(x)=0$?

I can show that for any $\varepsilon>0$, there is $\delta>0$ such that for any $0<a<b<\delta$, we have $|\int_a^b x^k g(x) dx|<\varepsilon$. However, the idea in If monotone decreasing and $\int_0^\infty f(x)dx <\infty$ then $\lim\limits_{x\to\infty} xf(x)=0.$ does not seem to apply here directly, because $x^\alpha g$ may not be monotone, but I also struggle to find a counterexample either. Is there anything that I'm missing?

score 2 · Accepted Answer · answered Mar 24 '23 at 19:15

Assume $g(x)\ge 0$ and $g$ is decreasing. Then $$\int\limits_0^xt^kg(t)\,dt\ge g(x)\int\limits_0^xt^k\,dt={1\over k+1}x^{k+1}g(x)\ge 0$$ Therefore $\displaystyle\lim_{x\to 0^+}x^{k+1}g(x)=0.$

If $g(x)\ge 0$ and $g$ is increasing then the conclusion holds obviously as $x^{k+1}\to 0$ and $g(x)$ has a finite limit when $x\to 0^+.$

Is it true that if $g$ is monotone on $(0,1]$ and $\int_0^1 x^k g$ exists for some $k>-1$, then $\displaystyle\lim_{x\to0} x^{k+1} g(x)=0$?

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