If $\Vert f \Vert_p < \infty$ for some $1 < p < \infty$ prove that $\lim_{t\to \infty} t^{-(1-1/p)} \int_0^t f(x) \, dx =0$.

Question

Question: Suppose $f \colon [0,\infty) \to \mathbb{R}$ is measurable so that $\Vert f \Vert_p < \infty$ for some $1 < p < \infty$. Prove that $$\lim_{t\to \infty} t^{-(1-1/p)} \int_0^t f(x) \, dx =0$$.

My attempt: I am able to show that $\limsup_{t \to \infty} t^{-(1-1/p)} \int_0^t f(x) \, dx \leq \Vert f \Vert_{p}$. Is there anyway I can continue with this problem?

Any hints are appreciated.

Answer 1

Below is a "functional analysis" approach.

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Let $q$ be the Hölder conjugate of $p$, i.e., $(1/p) + (1/q) = 1$. Let $g_n (x) := n^{-1/q} 1_{(0, n]} (x)$ for $x \in (0, \infty)$. Then $g_n \in L^q ((0, \infty); \mathbb R)$ with $\|g_n\|_q =1$ for $n \ge 1$. Clearly, $g_n \to 0$ everywhere. By Brezis' exercise 4.16.1, $g_n \to 0$ in the weak topology $\sigma (L^q, L^p)$. Then $\int g_n f \to 0$. The claim then follows.

Answer 2

The space $\mathcal{C}_{00}([0,\infty))$ is dense in $L_p$. Given $\varepsilon>0$ choose $\phi\in \mathcal{C}_{00}([0,\infty))$ such that $\|f-\phi\|_p<\varepsilon$. Then, by Holder's inequality, $$\int^t_0|f-\phi|\leq t^{1/q}\varepsilon$$ where $\frac1p+\frac1q=1$ (since $1<p<\infty$, so is $q$). Suppose $\operatorname{supp}(\phi)\subset[0,a]$. Then $$\int^t_0|\phi|\leq \|\phi\|_\infty a$$ for $t>a$. Putting things together $$t^{-1/q}\int^t_0|f|\leq t^{-1/q}\Big(\int^t_0|f-\phi|+\int^t_0|\phi|\Big)\leq \varepsilon+t^{-1/q}\|\phi\|_\infty a$$ Letting $t\rightarrow\infty$ yields $$\limsup_{t\rightarrow\infty}t^{-1/p}\int^t_0|f|\leq\varepsilon$$ The conclusion follows from this.