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timur
timur
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For $1\leq p\leq\infty$, the $r$-th order $L^p$-modulus of smoothness is \begin{equation} \omega_r(u,t,\Omega)_p=\sup_{|h|\leq t}\|\Delta_h^ru\|_{L^p(\Omega_{rh})} \end{equation} where $\Omega_{rh}=\{x\in\Omega:[x+rh]\subset\Omega\}$, and $\Delta_h^r$ is the $r$-th order forward difference operator defined recursively by $[\Delta_h^1u](x)=u(x+h)-u(x)$ and $\Delta_h^ku=\Delta_h^1(\Delta_h^{k-1})u$, i.e., \begin{equation} \Delta_h^ru (x) = \sum_{k=0}^r (-1)^{r+k} \binom{r}{k} u(x+kh). \end{equation} For $1\leq p,q\leq\infty$, $\alpha\geq0$, and $r>0$ an integer, let us define $B^\alpha_{p,q;r}(\Omega)$ as the space of functions $u\in L^p(\Omega)$ for which \begin{equation} |u|_{B^\alpha_{p,q;r}(\Omega)} = \big( \int_0^\infty|t^{-\alpha}\omega_r(u,t,\Omega)_p|^q \frac{\mathrm{d}t}t \big)^{1/q} < \infty, \end{equation} with the usual modifications for $q=\infty$. If $\alpha>r$ then the space $B^\alpha_{p,q;r}$ is trivial in the sense that $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$, the polynomials of order $r$. On the other hand, so long as $r>\alpha$, different choices of $r$ will result in norms that are equivalent to each other, and in this case we have the classical Besov spaces $B^\alpha_{p,q}(\Omega)=B^\alpha_{p,q;r}(\Omega)$. The reason for this is the so-called Marchaud inequalities.

My question concerns the borderline case $r=\alpha$. In this case, the situation seems to depend on the index $q$. Can you please confirm or invalidate the followings? Reference suggestions are also very welcome.

  • If $1\leq q<\infty$ and $\alpha=r$, then $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$.
  • The case $q=\infty$ gives nontrivial spaces. Namely, we have $B^\alpha_{p,\infty;\alpha}(\Omega)=W^{\alpha,p}(\Omega)$ for $p>1$, and $B^\alpha_{1,\infty;\alpha}(\Omega)$ consists of those functions whose derivatives of order up to $\alpha-1$ are in $BV(\Omega)$.

For the one dimensional case, the second item is proved on page 53 of Constructive Approximation by DeVore and Lorentz. I have not checked but suspect that the same proof works in multi-dimensions, with $BV(\Omega)=B^1_{1,\infty;1}(\Omega)$ now taken as a definition.

Now, what puzzles me is that on page 202 of Rational approximation of real functions by Petrushev and Popov, it is claimed that $B^\alpha_{p,p;\alpha}(\Omega)$ is nontrivial, which seems to contradict the first item above.

I am also interested in the full range $p,q>0$ involving quasi-Banach spaces. Here the borderline case would be given by $r=\alpha-\max\{0,\frac1p-1\}$.

For $1\leq p\leq\infty$, the $r$-th order $L^p$-modulus of smoothness is \begin{equation} \omega_r(u,t,\Omega)_p=\sup_{|h|\leq t}\|\Delta_h^ru\|_{L^p(\Omega_{rh})} \end{equation} where $\Omega_{rh}=\{x\in\Omega:[x+rh]\subset\Omega\}$, and $\Delta_h^r$ is the $r$-th order forward difference operator defined recursively by $[\Delta_h^1u](x)=u(x+h)-u(x)$ and $\Delta_h^ku=\Delta_h^1(\Delta_h^{k-1})u$, i.e., \begin{equation} \Delta_h^ru (x) = \sum_{k=0}^r (-1)^{r+k} \binom{r}{k} u(x+kh). \end{equation} For $1\leq p,q\leq\infty$, $\alpha\geq0$, and $r>0$ an integer, let us define $B^\alpha_{p,q;r}(\Omega)$ as the space of functions $u\in L^p(\Omega)$ for which \begin{equation} |u|_{B^\alpha_{p,q;r}(\Omega)} = \big( \int_0^\infty|t^{-\alpha}\omega_r(u,t,\Omega)_p|^q \frac{\mathrm{d}t}t \big)^{1/q} < \infty, \end{equation} with the usual modifications for $q=\infty$. If $\alpha>r$ then the space $B^\alpha_{p,q;r}$ is trivial in the sense that $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$, the polynomials of order $r$. On the other hand, so long as $r>\alpha$, different choices of $r$ will result in norms that are equivalent to each other, and in this case we have the classical Besov spaces $B^\alpha_{p,q}(\Omega)=B^\alpha_{p,q;r}(\Omega)$. The reason for this is the so-called Marchaud inequalities.

My question concerns the borderline case $r=\alpha$. In this case, the situation seems to depend on the index $q$. Can you please confirm or invalidate the followings?

  • If $1\leq q<\infty$ and $\alpha=r$, then $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$.
  • The case $q=\infty$ gives nontrivial spaces. Namely, we have $B^\alpha_{p,\infty;\alpha}(\Omega)=W^{\alpha,p}(\Omega)$ for $p>1$, and $B^\alpha_{1,\infty;\alpha}(\Omega)$ consists of those functions whose derivatives of order up to $\alpha-1$ are in $BV(\Omega)$.

For the one dimensional case, the second item is proved on page 53 of Constructive Approximation by DeVore and Lorentz. I have not checked but suspect that the same proof works in multi-dimensions, with $BV(\Omega)=B^1_{1,\infty;1}(\Omega)$ now taken as a definition.

Now, what puzzles me is that on page 202 of Rational approximation of real functions by Petrushev and Popov, it is claimed that $B^\alpha_{p,p;\alpha}(\Omega)$ is nontrivial, which seems to contradict the first item above.

I am also interested in the full range $p,q>0$ involving quasi-Banach spaces. Here the borderline case would be given by $r=\alpha-\max\{0,\frac1p-1\}$.

For $1\leq p\leq\infty$, the $r$-th order $L^p$-modulus of smoothness is \begin{equation} \omega_r(u,t,\Omega)_p=\sup_{|h|\leq t}\|\Delta_h^ru\|_{L^p(\Omega_{rh})} \end{equation} where $\Omega_{rh}=\{x\in\Omega:[x+rh]\subset\Omega\}$, and $\Delta_h^r$ is the $r$-th order forward difference operator defined recursively by $[\Delta_h^1u](x)=u(x+h)-u(x)$ and $\Delta_h^ku=\Delta_h^1(\Delta_h^{k-1})u$, i.e., \begin{equation} \Delta_h^ru (x) = \sum_{k=0}^r (-1)^{r+k} \binom{r}{k} u(x+kh). \end{equation} For $1\leq p,q\leq\infty$, $\alpha\geq0$, and $r>0$ an integer, let us define $B^\alpha_{p,q;r}(\Omega)$ as the space of functions $u\in L^p(\Omega)$ for which \begin{equation} |u|_{B^\alpha_{p,q;r}(\Omega)} = \big( \int_0^\infty|t^{-\alpha}\omega_r(u,t,\Omega)_p|^q \frac{\mathrm{d}t}t \big)^{1/q} < \infty, \end{equation} with the usual modifications for $q=\infty$. If $\alpha>r$ then the space $B^\alpha_{p,q;r}$ is trivial in the sense that $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$, the polynomials of order $r$. On the other hand, so long as $r>\alpha$, different choices of $r$ will result in norms that are equivalent to each other, and in this case we have the classical Besov spaces $B^\alpha_{p,q}(\Omega)=B^\alpha_{p,q;r}(\Omega)$. The reason for this is the so-called Marchaud inequalities.

My question concerns the borderline case $r=\alpha$. In this case, the situation seems to depend on the index $q$. Can you please confirm or invalidate the followings? Reference suggestions are also very welcome.

  • If $1\leq q<\infty$ and $\alpha=r$, then $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$.
  • The case $q=\infty$ gives nontrivial spaces. Namely, we have $B^\alpha_{p,\infty;\alpha}(\Omega)=W^{\alpha,p}(\Omega)$ for $p>1$, and $B^\alpha_{1,\infty;\alpha}(\Omega)$ consists of those functions whose derivatives of order up to $\alpha-1$ are in $BV(\Omega)$.

For the one dimensional case, the second item is proved on page 53 of Constructive Approximation by DeVore and Lorentz. I have not checked but suspect that the same proof works in multi-dimensions, with $BV(\Omega)=B^1_{1,\infty;1}(\Omega)$ now taken as a definition.

Now, what puzzles me is that on page 202 of Rational approximation of real functions by Petrushev and Popov, it is claimed that $B^\alpha_{p,p;\alpha}(\Omega)$ is nontrivial, which seems to contradict the first item above.

I am also interested in the full range $p,q>0$ involving quasi-Banach spaces. Here the borderline case would be given by $r=\alpha-\max\{0,\frac1p-1\}$.

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timur
timur
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Moduli of smoothness, Besov spaces, and Sobolev spaces

For $1\leq p\leq\infty$, the $r$-th order $L^p$-modulus of smoothness is \begin{equation} \omega_r(u,t,\Omega)_p=\sup_{|h|\leq t}\|\Delta_h^ru\|_{L^p(\Omega_{rh})} \end{equation} where $\Omega_{rh}=\{x\in\Omega:[x+rh]\subset\Omega\}$, and $\Delta_h^r$ is the $r$-th order forward difference operator defined recursively by $[\Delta_h^1u](x)=u(x+h)-u(x)$ and $\Delta_h^ku=\Delta_h^1(\Delta_h^{k-1})u$, i.e., \begin{equation} \Delta_h^ru (x) = \sum_{k=0}^r (-1)^{r+k} \binom{r}{k} u(x+kh). \end{equation} For $1\leq p,q\leq\infty$, $\alpha\geq0$, and $r>0$ an integer, let us define $B^\alpha_{p,q;r}(\Omega)$ as the space of functions $u\in L^p(\Omega)$ for which \begin{equation} |u|_{B^\alpha_{p,q;r}(\Omega)} = \big( \int_0^\infty|t^{-\alpha}\omega_r(u,t,\Omega)_p|^q \frac{\mathrm{d}t}t \big)^{1/q} < \infty, \end{equation} with the usual modifications for $q=\infty$. If $\alpha>r$ then the space $B^\alpha_{p,q;r}$ is trivial in the sense that $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$, the polynomials of order $r$. On the other hand, so long as $r>\alpha$, different choices of $r$ will result in norms that are equivalent to each other, and in this case we have the classical Besov spaces $B^\alpha_{p,q}(\Omega)=B^\alpha_{p,q;r}(\Omega)$. The reason for this is the so-called Marchaud inequalities.

My question concerns the borderline case $r=\alpha$. In this case, the situation seems to depend on the index $q$. Can you please confirm or invalidate the followings?

  • If $1\leq q<\infty$ and $\alpha=r$, then $B^\alpha_{p,q;r}=\mathbb{P}_{r-1}$.
  • The case $q=\infty$ gives nontrivial spaces. Namely, we have $B^\alpha_{p,\infty;\alpha}(\Omega)=W^{\alpha,p}(\Omega)$ for $p>1$, and $B^\alpha_{1,\infty;\alpha}(\Omega)$ consists of those functions whose derivatives of order up to $\alpha-1$ are in $BV(\Omega)$.

For the one dimensional case, the second item is proved on page 53 of Constructive Approximation by DeVore and Lorentz. I have not checked but suspect that the same proof works in multi-dimensions, with $BV(\Omega)=B^1_{1,\infty;1}(\Omega)$ now taken as a definition.

Now, what puzzles me is that on page 202 of Rational approximation of real functions by Petrushev and Popov, it is claimed that $B^\alpha_{p,p;\alpha}(\Omega)$ is nontrivial, which seems to contradict the first item above.

I am also interested in the full range $p,q>0$ involving quasi-Banach spaces. Here the borderline case would be given by $r=\alpha-\max\{0,\frac1p-1\}$.