# Welcome.

## January 19, 2011

### Harmonic Analysis Lecture Notes 7

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 6:50 pm

Weak-${L^p}$ and Lorentz spaces

For further information on this topic, see Chapter 1 of Grafakos’s text, “Classical Fourier Analysis”.

Definition For ${1 \leq p < \infty}$ and integer ${d \geq 1}$, we define the weak-${L^p({\mathbb R}^d)}$ space as the vector space of measurable functions ${f}$ on ${{\mathbb R}^d}$ such that

$\displaystyle ||f||^*_{L^{p,\text{weak}}({\mathbb R}^d)} = \sup_{0<\lambda<\infty} \lambda \cdot \big|\{ x : |f(x)| > \lambda \}\big|^{1/p} < \infty.$

Equivalently, the space consists of ${f}$ such that ${|\{ |f| > \lambda\}| \lesssim \lambda^{-p}}$.

Note: The “${*}$” is used in the notation to emphasize that this expression is not a norm (cf discussion on ${|| \:\:||^*_{L^{p,q}}}$).

For comparison’s sake, note that if ${f}$ is a measurable function on ${{\mathbb R}^d}$, then

$\displaystyle \begin{array}{rcl} ||f||_{L^p({\mathbb R}^d)} &=& \displaystyle \left( \int_{{\mathbb R}^d} \int_{0 \leq \lambda < |f(x)|} p\lambda^{p-1} \; d\lambda \; dx \right)^{1/p} \\ \\ &=& \displaystyle \left(\int_0^\infty \big|\{ |f| > \lambda \} \big| p\lambda^{p-1} \; d\lambda \right)^{1/p} \\ \\ &=& p^{1/p} \Big|\Big| \lambda |\{ |f| > \lambda \} |^{1/p} \Big|\Big|_{L^p\big( (0,\infty), \frac{d\lambda}{\lambda} \big)} \end{array}$

strongly resembles

$\displaystyle ||f||_{L^{p,\text{weak}}} = \underbrace{p^{1/\infty}}_{=1} \Big|\Big| \lambda |\{ |f| > \lambda\} |^{1/p} \Big|\Big|_{L^\infty\left( (0,\infty), \frac{d\lambda}{\lambda}\right)}$

and so suggests the following definition.

Definition For ${1 \leq p,q < \infty}$ and integer ${d \geq 1}$, we define the Lorentz space ${L^{p,q}({\mathbb R}^d)}$ as the vector space of measurable functions ${f}$ on ${{\mathbb R}^d}$ for which

$\displaystyle ||f||_{L^{p,q}}^* := p^{1/q} \Big|\Big| \lambda |\{ |f| > \lambda \}|^{1/p} \Big|\Big|_{L^q\left( (0,\infty), \frac{d\lambda}{\lambda}\right)}$

is finite. For the remainder of this discussion, we will write ${L^q\left(\frac{d\lambda}{\lambda}\right)}$ to mean ${L^q\left( (0,\infty), \frac{d\lambda}{\lambda}\right)}$.

Remark Indeed, ${L^{p,p} = L^p}$ and ${L^{p,\infty} = L^{p,\text{weak}}}$ by the above discussion.

Again ${|| \: \:\:||_{L^{p,q}}^*}$ is not a norm in general. Nevertheless, for all ${a \in {\mathbb C}}$,

$\displaystyle ||af||^*{L^{p,q}} = \Big|\Big| \lambda \big| \{ |f| > \frac{\lambda}{|a|} \} \big|^{1/p} \Big|\Big|_{L^q\left(\frac{d\lambda}{\lambda}\right)} = |a| \: \big|\big| f \big|\big|^*_{L^{p,q}}$

and

$\displaystyle \begin{array}{rcl} ||f+g||^*_{L^{p,q}} &=& \displaystyle \Big|\Big| \lambda \big|\{ |f+g| > \lambda\} \big|^{1/p} \Big|\Big|_{L^q\left(\frac{d\lambda}{\lambda}\right)}\\ &\leq& \displaystyle \Big|\Big| \lambda \left( \big| \{ |f| > \frac{\lambda}{2} \} \big| + \big| \{ |g| > \frac{\lambda}{2} \} \big| \right)^{1/p} \Big|\Big|_{L^q\left(\frac{d\lambda}{\lambda}\right)}\\ \\ \scriptsize\begin{bmatrix} \text{concavity of fractional}\\ \text{powers and Minkowski} \end{bmatrix} &\leq& \displaystyle \Big|\Big| \lambda \big| \{ |f| > \frac{\lambda}{2} \}\big|^{1/p} \Big|\Big|_{L^q\left(\frac{d\lambda}{\lambda}\right)} + \Big|\Big| \lambda \big| \{ |g| > \frac{\lambda}{2} \}\big|^{1/p} \Big|\Big|_{L^q\left(\frac{d\lambda}{\lambda}\right)} \\ \\ &\leq& 2 ||f||_{L^{p,q}}^* + 2||g||^*_{L^{p,q}}. \end{array}$

Thus, ${|| \:||_{L^{p,q}}^*}$ is a quasi-norm. When ${p \neq 1}$, this quasi-norm is equivalent to an actual norm, ${|| f||_{L^{p,q}}^* \lesssim_{p,q} || f||_{L^{p,q}} \lesssim_{p,q} || f||_{L^{p,q}}^*}$. When ${p=1}$ and ${q \neq 1}$, there cannot be a norm that is equivalent to ${|| \:||_{L^{p,q}}^*}$. Nevertheless, there is a metric which generates the same topology. In either case, we obtain a complete metric space.