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Hello again. Continuing with the book Stein, there is an observation that I would like to try.

I have the following:

For any $f\in L^1\cap L^2$, by proposition 1 in Bounded linear operators that commute with translation

$T(f)=f\ast \mu$ and by prop 2. in this post, $T(f)=\mathcal{F}^{-1}(m(x)\hat{f})$ then $\mathcal{F}^{-1}(m(x)\hat{f})=f\ast \mu$ implies $m(x)\hat{f}=F(f\ast\mu)=\hat{f}\hat{\mu}$. Therefore $\hat{f}(m-\hat{\mu})=0$ for any $f\in L^1\cap L^2$.

Therefore $m-\hat{\mu}=0$?

J. W. Tanner
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eraldcoil
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2 Answers2

1

Yes what are you not sure about. As usual look at $g_k = k^n e^{-\pi k^2 |x|^2}$ and $f\in C^\infty_c$, approximating $f_k=f\ast g_k$ with a sequence of finite sums of translates of $g_k$, since $T$ is continuous you get $$T(f_k) = T(g_k)\ast f,\qquad \widehat{T(f_k)} = \widehat{T(g_k)}\widehat{f}$$ by density this stays true for $f\in L^2$.

Taking $\widehat{f}= 1$ on $|x|<r$ you obtain your function $$m=\lim_{k\to \infty}\widehat{T(g_k)}\qquad (\text{convergence in } L^2_{loc})$$

It is (represented by) a measurable function because it is in $L^2_{loc}$.

This function is bounded because if the measure of $\{ x,|\widehat{T(g_k)}(x)| \ge A\}$ is non-zero then $\|T\|\ge A$.

reuns
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  • I did not understand the argument well. I was thinking of finding some suitable function f such that I can conclude that $m-\hat{\mu}=0$ – eraldcoil Jan 01 '20 at 00:22
  • In general, if $\hat{f}g=0$ for any $f\in L^1\cap L^2$ then $g=0$? proof: suppose $g(x_0)>0$ some $x_0$. I need a function $f\in L^1\cap L^2$ such that $\hat{f}(x_0)>0$ then $\hat{f}(x_0)g(x_0)>0$ a contradiction.

    What function works?

     – eraldcoil Jan 01 '20 at 01:03
  • If $g\neq 0$, exists $x_0\in\mathbb{R}^n$ such that $g(x_0)>0$. Let $\mathcal{X}{B(x_0,1)}$ characteristic function. Then $\hat{\mathcal{X}}{B(x_0,1)}(\xi)=\int_{B(x_0,1)}e^{ix\cdot \xi}dx>0$, in particular in $\xi=x_0$.Therefore, $\hat{\mathcal{X}}_{B(x_0,1)}(x_0)g(x_0)>0$ a contradiction. It is correct? – eraldcoil Jan 01 '20 at 01:15
  • No, you need to use the density, if a continuous operator vanishes on a dense subset then it vanishes everywhere. What do you not understand in my answer which is elementary ? That we can approximate the convolution $f\ast g_k$ by $\sum_n \frac1N f(n/N) g_k(x-n/N)$ whose image by $T$ is $\sum_n \frac1N f(n/N) T(g_k)(x-n/N)$ which approximates $f\ast T(g_k)$ ? – reuns Jan 01 '20 at 03:15
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A bounded linear operator $ T: L^2 \to L^2$ that commutes with translation

I will prove it for a Schwartz function $f$ ,you can easily extend the result for an $L^2$ function.

For a Schwartz function I will show :$f$ ,$T= f* \mu$ where $\mu $ is a tempered distribution.

A known result from Riesz Representation theorem see here is :$\int Tf(s)u(s) ds=\int f(s)T^*u(s)ds$ written as $<Tf,u>=<f,T^*u>$, where $ T^*$ is the linear adjoint operator of $ T $ , $ u \in L^2$

now for $f(s)$ translated by $x$ ,$f(s+x)$

we have $\int Tf(s+x)u(s)ds=\int f(s+x)T^*u(s)ds$

Now take Let $u_{\epsilon}=\phi_\epsilon(s)=\phi(s/\epsilon)\epsilon^{-1}$ where $\phi(s)$ is normalized gaussian function with zero mean.

$\int Tf(s+x)u_{\epsilon}(-s) ds=\int Tf(s+x)u_{\epsilon}(s)ds=\int f(s+x)T^*u_{\epsilon}(s)ds$

$H(s)=Tf(s)$ , by translation commutation $H(s+x)=Tf(s+x)$

$R(-s)=T^*u_{\epsilon}$

we have $\lim_{\epsilon \to 0} H* u_{\epsilon}=\lim_{\epsilon \to 0} f*R$

$\lim_{\epsilon \to 0} H*u_{\epsilon}=H(x)$ ae

for $u \in L^1$ function, we have the following theorem regarding fourier transform:

$v(x)=g*s=\int g(s)u(x-s) ds$ $L^2$ : for $g \in L^2$ define $\hat{g}=lim_{n \to \infty}\int_{n}^{-n}g(s)e^{2i\pi x z} dx$ the limit exists in the sense of $L^2$. If $u \in L^1$ and $\int |g(s)||u(x-s)| ds \le P$ where $P$ is a non negative integrable function in $L^2$ then $\hat{v}=\hat{g}\hat{u}$

Proof : $g_n=g1_{[-n,n]},v_n=g_n*u$ it's known that $\hat{v_n}=\hat{g_n}\hat{u}$ By dominated convergence theorem $lim_{n \to \infty} ||v-v_n||^2=0$ and by Plancherel theorem $lim_{n \to \infty} ||v-v_m||^2= lim_{n \to \infty} ||\hat{v}-\hat{v_n}||^2=0$ This implies $lim_{n \to \infty} \hat{v_n}=\hat{v}$

Now from our equation when $u$ is a normalized Gaussian we have $|H|* |u_{\epsilon}| \le MH(x)$ where $MH(x)$ is the Hardy Littlewood Maximal function (also $ ||MH(x)||^2 \le ||H(x)||^2$) using Fourier transform $\hat{H}(z)\hat{u_{\epsilon}}(z)=\hat{f}(z)G(z)$ where $G(z)=\hat{R}$

$\lim_{\epsilon \to 0}\hat{ u_{\epsilon}}=1$

$\lim_{\epsilon \to 0}G=m(z)$

consequently we have $\hat{Hf}(z)=m(z)\hat{f}(z)$

ibnAbu
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