Is $f(x) \in L^p(\mathbb R)$ always bounded for $x\longrightarrow\pm\infty$?

Question

I need to prove the following result on the derivative of an Hilbert transform for $f,f'\in L^p(\mathbb R)$

$$\mathcal H\bigg\{\frac{df(x)}{dx}\bigg\}=\frac{d}{dx}\mathcal Hf(x) $$

In particular from the right-hand side it follows (integrating by parts)

$$ \mathcal H\bigg\{\frac{df(x)}{dx}\bigg\}= \bigg[\frac{f(y)}{x-y}\bigg]_{\to-\infty}^{\to\infty}-\int...$$

then I have to prove that the first part of RHS is zero. That's why I need the result $$f\in L^p(\mathbb R)\Longrightarrow f\in L^{\infty}(\mathbb R)$$

Answer 1

That is not true in general, take e.g.

$$ f = \sum_n n \cdot \chi_{[n-2^{-n}, n + 2^{-n}]}. $$

But you always have $f(x_n) \to 0$ for a suitable sequence $x_n$ with $x_n \to \infty$ or $x_n \to -\infty$.

To see this, write

$$ \infty > \Vert f \Vert_p^p = \int_\Bbb{R} |f(x)|^p \,dx = \sum_{n \in \Bbb{Z}} \int_0^1 |f(x+n)|^p \,dx = \int_0^1 \sum_{n \in\Bbb{Z}} |f(x+n)|^p \,dx. $$

If the integral $\int |g(x)| \,dx$ is finite, this implies $|g(x)| < \infty$ for almost all $x$. Hence,

$$ |f(x + n)| \to 0 \text{ for } n \to \pm \infty \text{ for almost all } x \in [0,1]. $$

Answer 2

No, you could have a function of the form $$ \sum_n n \cdot \mathbb{1}_{[n, n + 1/2^n]}$$ which is in $L^p$ for all $0 <p < \infty$.