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Without loss of generality, $x=0$. Since $T(f_1,\dots,f_n)$ is $n$-linear in $(f_1,\dots,f_n)$, you can express each $f_i$ as a mixture of harmonics $\exp(it\cdot)$, $t\in\mathbb R$ (with, in general, complex coefficients), so that $T(f_1,\dots,f_n)$ is expressed as a mixture of the values $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ with real $t_1,\dots,t_n$. In turn, to express $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$, you can expand the product $\prod_{j=1}^n(e^{it_jx}-e^{it_jy})=\prod_{j=1}^n(1-e^{it_jy})$ in the numerator of the integrand in $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ and then proceed as in Appendix A in https://arxiv.org/abs/1603.07365, based on Lemma A.1 there (cf. Proposition A.4 there).

Addendum: What [Coifman et al.] seem to be doing concerning their formula (12) and what can be done in your situation seems really simple, much simpler than what was suggested above. For further simplicity, assume that $n=2$. Up to a constant factor, the Hilbert transform $H$ of a smooth enough function $g$ (say with a compact support) is given by the formula $$(Hg)(x)=\int\frac{g(x+t)-g(x)}t\,dt$$ for real $x$; the integrals here are over $\mathbb R$. So, differentiating in $x$ and then integrating by parts in $t$, one has $$(Hg)'(x)=\int\frac{g'(x+t)-g'(x)}t\,dt =\int\frac{g(x+t)-g(x)-g'(x)t}{t^2}\,dt.$$ Letting now $g(y):=(f_1(y)-f_1(0))(f_2(y)-f_2(0))$, one has $g(0)=g'(0)=0$, whence $$T(f_1,f_2)(0)=(Hg)'(0).$$ The general case of any natural $n$ is similar.

Without loss of generality, $x=0$. Since $T(f_1,\dots,f_n)$ is $n$-linear in $(f_1,\dots,f_n)$, you can express each $f_i$ as a mixture of harmonics $\exp(it\cdot)$, $t\in\mathbb R$ (with, in general, complex coefficients), so that $T(f_1,\dots,f_n)$ is expressed as a mixture of the values $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ with real $t_1,\dots,t_n$. In turn, to express $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$, you can expand the product $\prod_{j=1}^n(e^{it_jx}-e^{it_jy})=\prod_{j=1}^n(1-e^{it_jy})$ in the numerator of the integrand in $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ and then proceed as in Appendix A in https://arxiv.org/abs/1603.07365, based on Lemma A.1 there (cf. Proposition A.4 there).

Addendum: What [Coifman et al.] seem to be doing concerning their formula (12) and what can be done in your situation seems really simple, much simpler than what was suggested above. For further simplicity, assume that $n=2$. Up to a constant factor, the Hilbert transform $H$ of a smooth enough function $g$ (say with a compact support) is given by the formula $$(Hg)(x)=\int\frac{g(x+t)-g(x)}t\,dt$$ for real $x$; the integrals here are over $\mathbb R$ and understood in the principal value sense. So, differentiating in $x$ and then integrating by parts in $t$, one has $$(Hg)'(x)=\int\frac{g'(x+t)-g'(x)}t\,dt =\int\frac{g(x+t)-g(x)-g'(x)t}{t^2}\,dt.$$ Letting now $g(y):=(f_1(y)-f_1(0))(f_2(y)-f_2(0))$, one has $g(0)=g'(0)=0$, whence $$T(f_1,f_2)(0)=(Hg)'(0).$$ The general case of any natural $n$ is similar.

Without loss of generality, $x=0$. Since $T(f_1,\dots,f_n)$ is $n$-linear in $(f_1,\dots,f_n)$, you can express each $f_i$ as a mixture of harmonics $\exp(it\cdot)$, $t\in\mathbb R$ (with, in general, complex coefficients), so that $T(f_1,\dots,f_n)$ is expressed as a mixture of the values $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ with real $t_1,\dots,t_n$. In turn, to express $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$, you can expand the product $\prod_{j=1}^n(e^{it_jx}-e^{it_jy})=\prod_{j=1}^n(1-e^{it_jy})$ in the numerator of the integrand in $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ and then proceed as in Appendix A in https://arxiv.org/abs/1603.07365, based on Lemma A.1 there (cf. Proposition A.4 there).

What Coifman et al. seem to be doing concerning their formula (12) and what can be done in your situation seems really simple, much simpler than what was suggested above. For further simplicity, assume that $n=2$. Up to a constant factor, the Hilbert transform $H$ of a smooth enough function $g$ (say with a compact support) is given by the formula $$(Hg)(x)=\int\frac{g(x+t)-g(x)}t\,dt$$ for real $x$; the integrals here are over $\mathbb R$. So, differentiating in $x$ and then integrating by parts in $t$, one has $$(Hg)'(x)=\int\frac{g'(x+t)-g'(x)}t\,dt =\int\frac{g(x+t)-g(x)-g'(x)t}{t^2}\,dt.$$ Letting now $g(y):=(f_1(y)-f_1(0))(f_2(y)-f_2(0))$, one has $g(0)=g'(0)=0$, whence $$T(f_1,f_2)(0)=(Hg)'(0).$$ The general case of any natural $n$ is similar.

Without loss of generality, $x=0$. Since $T(f_1,\dots,f_n)$ is $n$-linear in $(f_1,\dots,f_n)$, you can express each $f_i$ as a mixture of harmonics $\exp(it\cdot)$, $t\in\mathbb R$ (with, in general, complex coefficients), so that $T(f_1,\dots,f_n)$ is expressed as a mixture of the values $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ with real $t_1,\dots,t_n$. In turn, to express $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$, you can expand the product $\prod_{j=1}^n(e^{it_jx}-e^{it_jy})=\prod_{j=1}^n(1-e^{it_jy})$ in the numerator of the integrand in $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ and then proceed as in Appendix A in https://arxiv.org/abs/1603.07365, based on Lemma A.1 there (cf. Proposition A.4 there).

Addendum: What [Coifman et al.] seem to be doing concerning their formula (12) and what can be done in your situation seems really simple, much simpler than what was suggested above. For further simplicity, assume that $n=2$. Up to a constant factor, the Hilbert transform $H$ of a smooth enough function $g$ (say with a compact support) is given by the formula $$(Hg)(x)=\int\frac{g(x+t)-g(x)}t\,dt$$ for real $x$; the integrals here are over $\mathbb R$. So, differentiating in $x$ and then integrating by parts in $t$, one has $$(Hg)'(x)=\int\frac{g'(x+t)-g'(x)}t\,dt =\int\frac{g(x+t)-g(x)-g'(x)t}{t^2}\,dt.$$ Letting now $g(y):=(f_1(y)-f_1(0))(f_2(y)-f_2(0))$, one has $g(0)=g'(0)=0$, whence $$T(f_1,f_2)(0)=(Hg)'(0).$$ The general case of any natural $n$ is similar.

Without loss of generality, $x=0$. Since $T(f_1,\dots,f_n)$ is $n$-linear in $(f_1,\dots,f_n)$, you can express each $f_i$ as a mixture of harmonics $\exp(it\cdot)$, $t\in\mathbb R$ (with, in general, complex coefficients), so that $T(f_1,\dots,f_n)$ is expressed as a mixture of the values $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ with real $t_1,\dots,t_n$. In turn, to express $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$, you can expand the product $\prod_{j=1}^n(e^{it_jx}-e^{it_jy})=\prod_{j=1}^n(1-e^{it_jy})$ in the numerator of the integrand in $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ and then proceed as in Appendix A in https://arxiv.org/abs/1603.07365, based on Lemma A.1 there (cf. Proposition A.4 there).

Without loss of generality, $x=0$. Since $T(f_1,\dots,f_n)$ is $n$-linear in $(f_1,\dots,f_n)$, you can express each $f_i$ as a mixture of harmonics $\exp(it\cdot)$, $t\in\mathbb R$ (with, in general, complex coefficients), so that $T(f_1,\dots,f_n)$ is expressed as a mixture of the values $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ with real $t_1,\dots,t_n$. In turn, to express $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$, you can expand the product $\prod_{j=1}^n(e^{it_jx}-e^{it_jy})=\prod_{j=1}^n(1-e^{it_jy})$ in the numerator of the integrand in $T(\exp(it_1\cdot),\dots,\exp(it_n\cdot))$ and then proceed as in Appendix A in https://arxiv.org/abs/1603.07365, based on Lemma A.1 there (cf. Proposition A.4 there).

What Coifman et al. seem to be doing concerning their formula (12) and what can be done in your situation seems really simple, much simpler than what was suggested above. For further simplicity, assume that $n=2$. Up to a constant factor, the Hilbert transform $H$ of a smooth enough function $g$ (say with a compact support) is given by the formula $$(Hg)(x)=\int\frac{g(x+t)-g(x)}t\,dt$$ for real $x$; the integrals here are over $\mathbb R$. So, differentiating in $x$ and then integrating by parts in $t$, one has $$(Hg)'(x)=\int\frac{g'(x+t)-g'(x)}t\,dt =\int\frac{g(x+t)-g(x)-g'(x)t}{t^2}\,dt.$$ Letting now $g(y):=(f_1(y)-f_1(0))(f_2(y)-f_2(0))$, one has $g(0)=g'(0)=0$, whence $$T(f_1,f_2)(0)=(Hg)'(0).$$ The general case of any natural $n$ is similar.