Let $p \in (1, \infty)$. If $f \in L^p(\mathbb{R})$ then $\lim_{x \rightarrow \infty} \int_{x}^{x+1}f(x)dx=0$

Question

Let $p \in (1, \infty)$. If $f \in L^p(\mathbb{R})$ then $$\lim_{x \rightarrow \infty} \int_{x}^{x+1}f(t)dt=0$$

Answer 1

• There are functions $f\in L^p(\Bbb R)$ such that $\limsup_{x\to\infty} f(x)=\infty$.

• There are functions $f\in L^p(\Bbb R)\setminus L^1(\Bbb R)$ for all $p>1$.

Therefore, no, this is not correct.

Added: Also, $\int_x^{x+1}f(t)\,dt\le f(x)$ needs not be true, but I'd say that this is the least important of the false assertions.

Answer 2

Hint:

$$\bigg|\int_x^{x+1} f(t)\,dt\bigg| \leq \int_x^{x+1} |f(t)|\,dt$$ $$\leq \bigg(\int_x^{x+1}|f(t)|^p\,dt\bigg)^{1 \over p}$$ The last inequality holds by Jensen's inequality.

So you just have to prove the statement when $p = 1$ since $|f(t)|^p \in L^1(\mathbb{R})$. Hint for this: Dominated Convergence.

Note I'm assuming the integral here is the integral of the function between $x$ and $x+1$ and not the integral of $f(x)$ since that doesn't make sense.