Rate of divergence of the integral of an $L^q$ function

Question

Let $p,q\in(1,\infty)$ be conjugate exponents (i.e., $1/p+1/q=1$) and let $f:(0,\infty)\to\mathbb R_+$ be a Lebesgue-measurable function such that $\int_0^{\infty} f(x)^q\,\mathrm dx<\infty$. It follows fairly easily from Hölder's inequality that $$\int_0^x f(t)\,\mathrm dt\leq x^{1/p}\cdot\|f\|_q\quad\forall x>0,$$ so that the (continuous) function $x\mapsto \int_0^x f(t)\,\mathrm dt$ must diverge no faster than $x^{1/p}$. (The notation $\int_0^x$ means Lebesgue integral on the set $\{t\in(0,\infty)\,|\,t<x\}$).

I want to show that the divergence of this partial integral is actually strictly slower than that of $x^{1/p}$, that is: $$\lim_{x\to\infty}x^{-1/p}\int_0^x f(t)\,\mathrm dt=0.$$ I put in considerable effort into trying to prove this, but none of my ideas worked. Any hint would be greatly appreciated.

Answer 1

It is not hard if $f$ is a linear combination of characteristic functions of sets of finite measure.

Then we use an approximation argument and the inequality $$\int_0^x|g(x)|\mathrm dt\leqslant x^{1/p}\lVert g\rVert_q$$ for any positive $x$ and $g\in L^q$.

Answer 2

Expanding on the answer of @DavideGiraudo:

Let $\varepsilon>0$ be arbitrary. The set of simple functions supported on finite-measure sets is dense in $L^q$, so there exists a function $g:(0,\infty)\to\mathbb C$ such that

• $g=\sum_{j=1}^{n}a_j\chi_{E_j}$, where $n$ is some positive integer;
• $a_j\in\mathbb C\setminus\{0\}$;
• $E_j$ are disjoint measurable sets;
• $m(E_j)<\infty$; and
• $\|f-g\|_q<\varepsilon$.

Let $x>0$. By the result mentioned in the question (via Hölder's inequality): $$\int_0^x|f(t)-g(t)|\,\mathrm dt\leq x^{1/p}\|f-g\|_q<x^{1/p}\varepsilon.$$

Now, putting these together, we have that \begin{align*} \int_0^x|f(t)|\,\mathrm dt\leq&\,\int_0^{x}|f(t)-g(t)|\,\mathrm dt+\int_0^{\infty}|g(t)|\,\mathrm dt<x^{1/p}\varepsilon+\int\left|\sum_{j=1}^n a_j\chi_{E_j}(t)\right|\,\mathrm dt\\ \leq&\,x^{1/p}\varepsilon+\int\sum_{j=1}^n|a_j|\chi_{E_j}(t)\,\mathrm dt=x^{1/p}\varepsilon+\underbrace{\sum_{j=1}^n|a_j|m(E_j)}_{\equiv A\in[0,\infty)}. \end{align*}

It follows that $$x^{-1/p}\int_0^x |f(t)|\,\mathrm dt<\varepsilon+x^{-1/p}A.$$ Hence, $$0\leq\limsup_{x\to\infty}\left\{x^{-1/p}\int_0^x|f(t)|\,\mathrm dt\right\}\leq\varepsilon.$$ This is true for all $\varepsilon>0$, so $\limsup =0$. Also, $\liminf\geq0$ because of non-negativity. It follows that $\limsup=\liminf=\lim=0$, as required.