4 Presumeably the OP meant d/dx, else the LHS is just zero trivially. ; edited tags

hi,

assume that I have a function $q$ which is a Fourier Multiplier of order zero, i.e. $$\left|\left( \frac{d}{dt}\right)^nq(x)\right|\lesssim frac{d}{dx}\right)^nq(x)\right|\lesssim \left(\frac{1}{1+|x|}\right)^n\quad \mbox{for all }n\geq 0.$$ Can I conclude that $q$ is the Fourier Transform of a finite Radon Measure? If not, what are the conditions on $q$?

Actually, what I really want to know is when is the Fourier Multiplication Operator defined by $q$ bounded on $L^p$? The Mihlin Multiplier theorem gives an affirmative answer for $1<p<\infty$. How about the boundary cases $1$ and $\infty$?

3 backticks to fix math

hi,

assume that I have a function $q$ which is a Fourier Multiplier of order zero, i.e. $$\left|\left( \frac{d}{dt}\right)^nq(x)\right|\lesssim \left(\frac{1}{1+|x|}\right)^n\quad \mbox{for all }n\geq 0.$$ Can I conclude that $q$ is the Fourier Transform of a finite Radon Measure? If not, what are the conditions on $q$?

Actually, what I really want to know is when is the Fourier Multiplication Operator defined by $q$ bounded on $L^p$? The Mihlin Multiplier theorem gives an affirmative answer for $1<p<\infty$. How about the boundary cases $1$ and $\infty$?

2 added 245 characters in body

hi,

assume that I have a function $q$ which is a Fourier Multiplier of order zero, i.e. $$\left|\left( \frac{d}{dt}\right)^nq(x)\right|\lesssim \left(\frac{1}{1+|x|}\right)^n\quad \mbox{for all }n\geq 0.$$ Can I conclude that $q$ is the Fourier Transform of a finite Radon Measure? If not, what are the conditions on $q$?

Actually, what I really want to know is when is the Fourier Multiplication Operator defined by $q$ bounded on $L^p$? The Mihlin Multiplier theorem gives an affirmative answer for \$1

1