Let $\phi,\psi$ be functions on $\mathbb{R}^d$ with $\int\phi=\int\psi=0$ and $$|\phi(x)|,|\psi(x)|\lesssim(1+|x|)^{-(d+1)} \quad\text{and}\quad |\nabla\phi(x)|,|\nabla\psi(x)|\lesssim(1+|x|)^{-(d+2)}. $$ Let $\phi_t(x)=t^{-d}\phi(x/t)$ and similarly for $\psi$. For $R>1$ define the operator $$T_R f:=\int_{1/R}^R\int_{\mathbb{R}^d}(f*\psi_t)(y) \phi_t(\cdot-y)\mathrm{d}y\frac{\mathrm{d}t}{t}. $$
I have to show that $T_R$ is uniformly bounded in $L^2(\mathbb R^d)$ for $R>1$. There is a hint which suggests to use the duality of $L^2$ and the almost orthogonality lemma which says the following:
Let $\varphi:\mathbb R^d\rightarrow\mathbb R$ be a function such that for all $z,z'\in\mathbb R^d$ we have:
1. $\int_{\mathbb R^d}\varphi=0$
2. $|\varphi(z)|\leq(1+|z|)^{-(d+1)}$
3. $|\varphi(z)-\varphi'(z)|\leq|z-z'|((1+|z|)^{-(d+1)}+(1+|z'|)^{-(d+1)})$.
Let $\varphi_{y,t}(z)=t^{-d}\varphi(t^{-1}(y-z))$. Then $$|\langle\varphi_{y,t},\varphi_{y',t'}\rangle|\leq\frac{tt'}{\max(t,t')+|y-y'|^{d+1}}.$$
So I've gotten to this estimate $$T_R f(x)\lesssim \int_{1/R}^R |\langle f*\varphi_t,\varphi_{x,t} \rangle|\frac{\mathrm{d}t}{t} \lesssim\int_{1/R}^R\frac{t}{t+|x|^{d+1}}\frac{\mathrm{d}t}{t} \leq\int_{1/R}^R\frac{1}{t}\mathrm{d}t =2\ln(R),$$ but I feel like it is not good enough, since it does not give the independence on $R>1$.
In part two it asks to show that $\lim_{R,R'\rightarrow\infty}\|T_R f-T_{R'}f\|_{L^2}=0$.
I imagine it follows from the uniform bound, but I suspect I have to do the estimate of the above first so that i can estimate the limit and then pass to the second limit.
EDIT: I think I got a decent attempt. Let $\mathcal F:g\mapsto\widehat{g}$ denote the Fourier transform operator and let $\phi_{y,t}:=\phi_t(y-x)$. Then $$|\langle T_R f,g \rangle| \leq\int_{\mathbb{R}^d}g(x)\int_{1/R}^R\int_{\mathbb{R}^d} (f*\psi_t)(y)\phi_t(x-y) \mathrm{d}y\frac{\mathrm{d}t}{t}\mathrm{d}x $$ $$\overset{\text{Fubini}}=\int_{1/R}^R \int_{\mathbb{R}^d}(f*\psi_t)(y) \int_{\mathbb{R}^d}g(x)\phi_t(x-y) \mathrm{d}x\mathrm{d}y\frac{\mathrm{d}t}{t} $$ $$=\int_{1/R}^R \langle f*\psi_t,g*\phi_{0,t} \rangle \frac{\mathrm{d}t}{t} \overset{\text{Plancherel}}=\int_{1/R}^R \langle \mathcal{F}(f*\psi_t),\mathcal{F}(g*\phi_{0,t})\rangle \frac{\mathrm{d}t}{t} $$ $$=\int_{1/R}^R \langle\widehat f\cdot\widehat\psi_t, \widehat g\cdot\widehat\phi_{0,t}) \rangle \frac{\mathrm{d}t}{t} $$ $$\leq C|\langle f,g \rangle| \int_{1/R}^R\langle\psi_t,\phi_{0,t}\rangle\frac{\mathrm{d}t}{t} $$ $$\overset{\text{lemma from above??}}\leq C'|\langle f,g \rangle| $$ I feel like this last step does not work. However, if it does work, then choosing $g=T_R f$ gives $$\|T_R f\|_{L^2} \leq\frac{C'}{\|T_R f\|_{L^2}}|\left\langle f,T_R f \right\rangle| \overset{\text{Cauchy-Schwarz}}\leq C'\frac{\|T_R f\|_{L^2}}{\|T_R f\|_{L^2}}\|f\|_{L^2} =C'\|f\|_{L^2}$$ for all $R>1$.
For part 2 I had the following idea: Let $1<R'<R$. Then \begin{align*} |\langle T_R f-T_{R'} f,g \rangle| &\overset{\text{part (a)}}\leq C|\langle f,g\rangle| \underbrace{\left( \int_{1/R}^{1/R'} |\langle \psi_t,\phi_{0,t} \rangle| \frac{\mathrm{d}t}{t} +\int_{R'}^R |\langle \psi_t,\phi_{0,t} \rangle| \frac{\mathrm{d}t}{t} \right)}_{\text{bounded independently of }R>R'} \end{align*} Therefore $$\lim_{R'\rightarrow\infty}\lim_{R\rightarrow\infty}|\langle T_R f-T_{R'} f,g \rangle| \leq\lim_{R'\rightarrow\infty} C|\langle f,g\rangle| \left( \int_0^{1/R'} |\langle \psi_t,\phi_{0,t} \rangle| \frac{\mathrm{d}t}{t} +\int_{R'}^\infty |\langle \psi_t,\phi_{0,t} \rangle| \frac{\mathrm{d}t}{t} \right) =0$$ Hence $\lim_{R,R'\rightarrow\infty}\|T_R f-T_{R'}f\|_{L^2}=0$.