A bounded linear operator $ T: L^2 \to L^2$ with these properties :
- Commutes with translation
- Commutes with dilation
- Has in its kernel functions $ f $ such that support $\hat {f} \subseteq [0,\infty) $. where $\hat {f}$ is the fourier transform of $f$
For a Schwartz function$f$ ,$T= f* \mu$ where $\mu $ is a tempered distribution.
Then $\int_{-\infty}^0\hat {f}(s)e^{2i\pi s x} ds =c Tf (x)$ where $ c $ is s constant. Is the proof below correct ?
Known result from Riesz Representation theorem see here:$\int Tf(s)u(s) ds=\int f(s)T^*u(s)ds$ 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:
$v(x)=g*u=\int g(s)u(x-s) ds$ 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 $\hat{Hf}(z)=m(z)\hat{f}(z)$
Therefore $L^2$ limit: $H(x)=\lim_{n \to \infty}\int_{-n}^{n}m(z)\hat {f}(z)e^{2i\pi x z} dz$
Now using dilation commutation $\lim_{n \to \infty}\int_{-n}^{n}m(z)\hat {f}(z)e^{2i\pi s z} dz=\lim_{n \to \infty}\int_{-n}^{n}a^{-1}m(z)\hat {f}(a^{-1}z)e^{2i\pi a^{-1}s z} dz=\lim_{n \to \infty}\int_{-a^{-1}n}^{a^{-1}n}m(az)\hat {f}(z)e^{2i\pi s z} dz$
define $ g=m(az)\hat{f}(z)$
define $g_n=g_{[{-n,n}]}-g_{[{-a^{-1}n,a^{-1}n}]}$
$\lim_{n \to \infty} ||g_n||^2=0$
$\hat{g_n}=\int_{-n}^{n}m(az)\hat {f}(z)e^{2i\pi s z} dz-\int_{-a^{-1}n}^{a^{-1}n}m(az)\hat {f}(z)e^{2i\pi s z} dz$
By Plancherel theorem $\lim_{n \to \infty} ||\hat{g_n}||^2=\lim_{n \to \infty} ||g_n||^2=0$
Therefore there is a subsequence $\{n\}$ such that $\lim_{n \to \infty}\int_{-a^{-1}n}^{a^{-1}n}m(az)\hat {f}(z)e^{2i\pi s z} = \lim_{n \to \infty}\int_{-n}^{n}m(az)\hat {f}(z)e^{2i\pi s z}$
so $\lim_{n \to \infty}\int_{-n}^{n}(m(z)-m(az))\hat {f}(z)e^{2i\pi s z} dz=0$ and by using Plancherel theorem this implies $m(az)=m(z)$ ae for any $a>0$
Therefore $m(z)=c_{-} $ ae on $z \in (-\infty ,0)$ and $m(z)=c_{+}$ ae on $ z \in (0,\infty)$ according to see here
Given the 3rd condition $c_{+}\int_{0}^{\infty}\hat {f}(z)e^{2i\pi x z}dz=0$,since we know $\int_{0}^{\infty}\hat {f}(z)e^{2i\pi x z}dz\ne 0$ ,so $c_{+}=0$
Therefore $Hf(x)=c_{-}\int_{-\infty}^0\hat {f}(z)e^{2i\pi z x} dz$ ae
