Skip to main content

Hello,

the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in \[1,\infty\]$$p \in [1,\infty]$ are measurable functions with certain decay properties at infinity or at the singularities.

In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function.

In contrast to that, hardy spaces seem to be less intuitive due to cancellation properties. Hence I wonder:

  1. How do "typical" $H^1$ functions look like?
  2. In particular, what can you typically "do" with Hardy functions?

For example, I guess convolution arguments take a more prominent role in approximation arguments than perturbation arguments do, but I am not sure about that.

The hardy space $H^1$ shall be defined via

$f \in H^1 $ if $f \in L^1_{\mathrm{loc}}$ and $Mf \in L^1$, where

$$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$

$$\mathcal{L}(B_r(x_0)) = \left\{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0|,0)}{r|B_r(x_0)|} , \mathrm{Lip}(\phi) < \frac{1}{r|B_r(x_0)|} \right\}$$

Hello,

the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in \[1,\infty\]$ are measurable functions with certain decay properties at infinity or at the singularities.

In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function.

In contrast to that, hardy spaces seem to be less intuitive due to cancellation properties. Hence I wonder:

  1. How do "typical" $H^1$ functions look like?
  2. In particular, what can you typically "do" with Hardy functions?

For example, I guess convolution arguments take a more prominent role in approximation arguments than perturbation arguments do, but I am not sure about that.

The hardy space $H^1$ shall be defined via

$f \in H^1 $ if $f \in L^1_{\mathrm{loc}}$ and $Mf \in L^1$, where

$$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$

$$\mathcal{L}(B_r(x_0)) = \left\{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0|,0)}{r|B_r(x_0)|} , \mathrm{Lip}(\phi) < \frac{1}{r|B_r(x_0)|} \right\}$$

the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in [1,\infty]$ are measurable functions with certain decay properties at infinity or at the singularities.

In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function.

In contrast to that, hardy spaces seem to be less intuitive due to cancellation properties. Hence I wonder:

  1. How do "typical" $H^1$ functions look like?
  2. In particular, what can you typically "do" with Hardy functions?

For example, I guess convolution arguments take a more prominent role in approximation arguments than perturbation arguments do, but I am not sure about that.

The hardy space $H^1$ shall be defined via

$f \in H^1 $ if $f \in L^1_{\mathrm{loc}}$ and $Mf \in L^1$, where

$$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$

$$\mathcal{L}(B_r(x_0)) = \left\{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0|,0)}{r|B_r(x_0)|} , \mathrm{Lip}(\phi) < \frac{1}{r|B_r(x_0)|} \right\}$$

fixed spelling error, edited math display; edited body; edited tags
Source Link
Willie Wong
  • 39.1k
  • 4
  • 94
  • 176

Hello,

the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in \[1,\infty\]$ are measurable functions with certain decay properties at infinity or at the singularities.

In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function.

In contrast to that, hardy spaces seem to be less intuitive due to cancelationcancellation properties. Hence I wonder:

  1. How do "typical" $H^1$ functions look like?
  2. In particular, what can you typically "do" with Hardy functions?

For example, I guess convolution arguments take a more prominent role in approximation arguments than pertubationperturbation arguments do, but I am not sure about that.

The hardy space $H^1$ shall be defined via

$f \in H^1 $ if $f \in L^1_{loc}$$f \in L^1_{\mathrm{loc}}$ and $Mf \in L^1$, where

$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$

$L(B_r(x_0)) = \{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0,|0)}{r|B_r(x_0)|} , Lip(\phi) < \frac{1}{r|B_r(x_0)|} \}$$$\mathcal{L}(B_r(x_0)) = \left\{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0|,0)}{r|B_r(x_0)|} , \mathrm{Lip}(\phi) < \frac{1}{r|B_r(x_0)|} \right\}$$

Hello,

the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in \[1,\infty\]$ are measurable functions with certain decay properties at infinity or at the singularities.

In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function.

In contrast to that, hardy spaces seem to be less intuitive due to cancelation properties. Hence I wonder:

  1. How do "typical" $H^1$ functions look like?
  2. In particular, what can you typically "do" with Hardy functions?

For example, I guess convolution arguments take a more prominent role in approximation arguments than pertubation arguments do, but I am not sure about that.

The hardy space $H^1$ shall be defined via

$f \in H^1 $ if $f \in L^1_{loc}$ and $Mf \in L^1$, where

$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$

$L(B_r(x_0)) = \{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0,|0)}{r|B_r(x_0)|} , Lip(\phi) < \frac{1}{r|B_r(x_0)|} \}$

Hello,

the standard intuition for Lebesgue spaces $L^p(\mathbb R^n)$ for $p \in \[1,\infty\]$ are measurable functions with certain decay properties at infinity or at the singularities.

In particular, a typical $L^p$ function is a function like $|x|^{-\alpha}$ with $\alpha > \frac{n}{p}$ for $p$ < $\infty$, or simply $1$ for $p = \infty$. Furthermore, functions which are almost everywhere absolute-value-dominated by an $L^p$ function are elements of $L^p$, too. This is useful for approximation arguments, as the pointwise error has just to be dominated by an $\epsilon$-multiple of another $L^p$ function.

In contrast to that, hardy spaces seem to be less intuitive due to cancellation properties. Hence I wonder:

  1. How do "typical" $H^1$ functions look like?
  2. In particular, what can you typically "do" with Hardy functions?

For example, I guess convolution arguments take a more prominent role in approximation arguments than perturbation arguments do, but I am not sure about that.

The hardy space $H^1$ shall be defined via

$f \in H^1 $ if $f \in L^1_{\mathrm{loc}}$ and $Mf \in L^1$, where

$$Mf(x) := \sup \limits_{B_r(x_0) \ni x, \phi \in \mathcal L(B_r(x_0))} \int f \phi dx$$

$$\mathcal{L}(B_r(x_0)) = \left\{ \phi \in C(B_r(x_0)) s.t. |\phi(x)| < \dfrac{\max(r-|x-x_0|,0)}{r|B_r(x_0)|} , \mathrm{Lip}(\phi) < \frac{1}{r|B_r(x_0)|} \right\}$$

tag added
Link
shuhalo
  • 5.3k
  • 29
  • 43
Source Link
shuhalo
  • 5.3k
  • 29
  • 43
Loading