Show that F vanishes at infinity.

Question

Suppose $1 ≤ p < ∞, f ∈ L^p(R)$, and

$F(x) = \int_{x}^{x+1} f(t) dm(t)$

Prove that F vanishes at infinity.

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We know that $\int_R |f|^p < \infty$, then, of course, for any $x, F(x)< \int_x^{x+1} |f|^p < \infty$. But I want to show that not only is it finite, but it goes to zero as $x$ goes to $\infty$.

I was thinking to approach it by way of contradiction. Assume that,

(WLOG assume $f \geq 0$)

$lim_{x \rightarrow \infty} F(x) > 0$

$\implies lim_{x \rightarrow \infty} \int_x^{x+1} f(t) dm(t) > 0$

I thought perhaps to approach this with partial sums.. and show that it contradicts that $f \in L^p(R)$. But I have had no luck with this.

Help? Hints? I would greatly appreciate it!

Answer 1

Let $q$ be the conjugated exponent of $p$.

Hölder inequality on $[x,x+1]$ and Lebesgue dominated convergence theorem imply \begin{align} |F(x)| &\le \int_{x}^{x+1} |f| \\ &\le \left(\int_x^{x+1} |f|^p\right)^{1/p}\left(\int_x^{x+1} 1\right)^{1/q} \\ &=\left(\int_{\langle -\infty, x+1]}|f|^p - \int_{\langle -\infty, x]}|f|^p\right)^{1/p} \\&\xrightarrow{x\to\infty} \left(\|f\|_p^p - \|f\|_p^p\right)^{1/p} \\ &= 0 \end{align} so $\lim_{x\to\infty} F(x) = 0$.

Answer 2

Let $g_n(t) = |f(t)|^p\cdot \chi_{[-n,n]}(t)$. Note that $0\leqslant g_n(t)\nearrow |f(t)|^p$ pointwise a.e. as $n\to\infty$. By the Monotone Convergence Theorem, $$ \int g_n(t)\,dt \to \int |f(t)|^p\,dt \quad\text{as $n\to\infty$}. $$ Hence the difference $$ \int|f(t)|^p\,dt - \int_{-n}^n|f(t)|^p\,dt \geqslant \int_n^\infty|f(t)|^p\,dt $$ can be made as small as desired by taking $n$ sufficiently large. In particular, the integral $$ \int_x^\infty|f(t)|^p\,dt \geqslant \int_x^{x+1}|f(t)|^p\,dt $$ can be made arbitrarily small by taking $x$ sufficiently large. If $p = 1$, then we are done by the triangle inequality applied to $F(x)$.

If $p > 1$, apply Hölder's inequality \begin{align*} |F(x)| &\leqslant \int_x^{x+1}|f(t)|\,dt \leqslant \bigg(\int_{x}^{x+1}|f(t)|^p\,dt\bigg)^{1/p}\bigg(\int_{x}^{x+1}1^q\,dt\bigg)^{1/q}=\bigg(\int_{x}^{x+1}|f(t)|^p\,dt\bigg)^{1/p}, \end{align*} and the estimate from above.

Answer 3

For $p=1$ this is clear since $\int_x^{\infty} |f| \to 0$. For $p>1$ we have $|\int_x^{x+1} f|=|\int_x^{x+1} (1)f|\leq (\int_x^{x+1}|f|^{p})^{1/p}(\int 1^{q})^{1/q}\leq \int_x^{\infty} |f|^{p} \to 0$. [Here $q=\frac p {p-1}$].

Answer 4

$$ \|f\|_p^p \ge\sum_{k=0}^\infty\int_k^{k+1}|f(x)|^p\,\mathrm{d}x\\ $$ Since $f\in L^p(\mathbb{R})$, the sum converges; therefore, the terms must tend to $0$: $$ \begin{align} \lim_{k\to\infty}\int_k^{k+1}|f(x)|\,\mathrm{d}x &\le\lim_{k\to\infty}\left(\int_k^{k+1}|f(x)|^p\,\mathrm{d}x\right)^{1/p}\\[6pt] &=0 \end{align} $$ Thus, $$ \begin{align} \lim_{x\to\infty}|F(x)| &\le\lim_{x\to\infty}\int_{\lfloor x\rfloor}^{\lfloor x\rfloor+1}|f(x)|\,\mathrm{d}x+\lim_{x\to\infty}\int_{\lfloor x+1\rfloor}^{\lfloor x\rfloor+2}|f(x)|\,\mathrm{d}x\\[6pt] &=0 \end{align} $$