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In Bounded linear operators that commute with translation

It has: Let $m(\xi)=(T^{\;*}\psi)^\wedge(\xi)/\psi(\xi)$, then we have $$ (Tf)^\wedge(\xi)=m(\xi)\hat{f}(\xi) $$ and therefore, $\|m\|_{L^\infty}=\|T\|_{L^2}$.

Why $\|m\|_{L^\infty}=\|T\|_{L^2}$. I can't see this.

Marios Gretsas
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eraldcoil
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1 Answers1

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Hint: $\|T\|_{L^{2}} \leq \|m\|_{\infty}$ is obvious from the definition of the norm. Now let $N$ be a positive integer and an $\epsilon >0$ and consider $I_A$ where $A=\{\xi: |m(\xi)| > \|m\|_{\infty} -\epsilon, \|x\|\leq N\}$. (We can choose $N$ such that $A$ has positive measure). This is an $L^{2}$ function and hence it can be written as $\hat {f}$ for some $L^{2}$ function $f$. Now $\|T\|_{L^{2}} \geq \frac {\|m(\xi)\hat {f}(\xi)\|_2} {\|f(\xi)\|_2} $ by definition of the norm of $T$. Now just simplify the last expression and let $\epsilon \to 0$.

Kavi Rama Murthy
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  • I did not understand very well how to continue. $\frac {| m(\xi)\hat {f}(\xi)|2} {|f(\xi)|_2} =\frac{| m(\xi) m(\xi)I{A} |{2}}{| \hat{f}|_2} =\frac{| m(\xi) m(\xi)I{A} |2}{| m(\xi)I{A}|_{2}}$ – eraldcoil Dec 26 '19 at 05:59
  • @eraldcoil My answer was wrong. I have corrected it now. – Kavi Rama Murthy Dec 26 '19 at 06:12
  • i have doubts. $\frac{| m(\xi) m(\xi)I_{A} |2}{| m(\xi)I{A}|{2}}=\frac{\int{A}m^2(x)dx}{\int_{A} |m(x)|dx}...$ – eraldcoil Dec 26 '19 at 06:26
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    On $A$ we have $|m(\xi)| >|m|{\infty} -\epsilon$. So $\int_A |m(\xi)|^{2} \geq (|m|{\infty} -\epsilon)^{2} \lambda (A)$ where $\lambda$ is the Lebesgue measure. Note that $\sqrt {\lambda(A)}$ cancels when you take the ratio. – Kavi Rama Murthy Dec 26 '19 at 06:30
  • $|T|^2\geq$ $\frac{| m(\xi) m(\xi)I_{A} |2^2}{| m(\xi)I{A}|{2}^2} =\frac{ \int{A} |m(x)|^4dx}{\int_{A}|m(x)|^2dx} \geq \frac{(|m|{\infty}-\epsilon)^4\lambda(A)}{|m|{\infty}^2\lambda(A)} =\frac{(|m|{\infty}-\epsilon)^4}{|m|{\infty}^2} \to |m|_{\infty}^2\$

    when $\epsilon\to 0$

    Therefore $|T|\geq |m|_{\infty}$ !! However, can't see the need for $N \to \infty$

     – eraldcoil Dec 26 '19 at 07:01
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    No need to let $N \to \infty$. But for the argument to be valid yo u have to make sure that you don't get $0$ in the denominator. If $N$ is large then $\lambda (A) >0$. You can fix any $N$ with this property. @eraldcoil – Kavi Rama Murthy Dec 26 '19 at 07:08
  • I have a doubt.

    Why is inequality $|T|\leq |m|{\infty} $obvious? I have this: $|T|=\sup{|f|{2}\leq 1}|T(f)|{2}\leq \sup_{|f|{2}\leq 1} |m|{2}$

     – eraldcoil Jan 13 '20 at 00:08
  • @eraldcoil Recall that $f$ and $\hat f$ have the same $L^{2}$ norm. – Kavi Rama Murthy Jan 13 '20 at 00:10
  • thanks. $|(Tf)^{\wedge}|{2}=|m\hat{f}|{2}\leq |m|{\infty} |\hat{f}|{2}$ Plancherel implies $|T(f)|{2}\leq |m|{\infty}|f|{2}$ then $\sup{|f|\leq 1} |T(f)|{2}\leq |m|{\infty}$ i.e. $|T|\leq |m|_{\infty}$! – eraldcoil Jan 13 '20 at 00:18