Denis Serre's answer is just perfect. Let me add a couple of examples of distributions that can be squared:

(1) With $H$ the Heaviside function, define $\operatorname{Log}(x+i0)=\ln(\vert x\vert)+i\pi H(-x) $ and $$T_1=\frac{1}{x+i0}=\frac{d}{dx}(\operatorname{Log}(x+i0)) = \operatorname{pv}\frac 1{x}-i\pi \delta_0(x).$$ It is easy to see that $ WF T_1=[0]\times (0,+\infty), $ so that $WF T_1+WF T_1$ does not meet $0.$ Then there is no difficulty to define $T^2$ say as $$ \langle T^2,\phi\rangle=\lim_{\varepsilon\rightarrow 0_+} \int\frac{\varphi(x) \, dx}{(x+i\varepsilon)^2}. $$

(2) Let us consider a smooth hypersurface $\Sigma$ of $\mathbf R^d$ defined by the equation $f(x)=0$ with a smooth $f$ such that $df\not=0$ at $f=0$ and let $\delta_\Sigma$ be the Euclidean measure on $\Sigma$. Then $$ T_2=pv\frac{1}{f}-i\delta_\Sigma $$ can be squared. The reason is the same than for the previous example, since $ WF T_2$ is the positive conormal of $\Sigma$. A point $(x,\xi)\in WF T_2$ iff $$ x\in \Sigma\quad \xi =\lambda df(x) \text{ with $\lambda >0$}. $$ Then of course, if $(x,\xi_j)$, $j=1,2$ are both in $WF T_2$ then $$ \xi_1+\xi_2\not=0. $$

Denis Serre's answer is just perfect. Let me add a couple of examples of distributions that can be squared:

1. With $H$ the Heaviside function, define $\operatorname{Log}(x+i0)=\ln(\vert x\vert)+i\pi H(-x) $ and $$T_1=\frac{1}{x+i0}=\frac{d}{dx}(\operatorname{Log}(x+i0)) = \operatorname{pv}\frac 1{x}-i\pi \delta_0(x).$$ It is easy to see that $ \DeclareMathOperator\WF{WF}\WF T_1=[0]\times (0,+\infty), $ so that $\WF T_1+\WF T_1$ does not meet $0$. Then there is no difficulty to define $T^2$ say as $$ \langle T^2,\phi\rangle=\lim_{\varepsilon\rightarrow 0_+} \int\frac{\varphi(x) \, dx}{(x+i\varepsilon)^2}. $$

2. Let us consider a smooth hypersurface $\Sigma$ of $\mathbf R^d$ defined by the equation $f(x)=0$ with a smooth $f$ such that $df\ne0$ at $f=0$ and let $\delta_\Sigma$ be the Euclidean measure on $\Sigma$. Then $$ T_2=pv\frac{1}{f}-i\delta_\Sigma $$ can be squared. The reason is the same as for the previous example, since $ \WF T_2$ is the positive conormal of $\Sigma$. A point $(x,\xi)\in \WF T_2$ iff $$ x\in \Sigma\quad \xi =\lambda df(x) \text{ with $\lambda >0$}. $$ Then of course, if $(x,\xi_j)$, $j=1,2$ are both in $\WF T_2$ then $$ \xi_1+\xi_2\ne0. $$

Denis Serre's answer is just perfect. Let me add a couple of examples of distributions that can be squared:

(1) With $H$ the Heaviside function, define $Log(x+i0)=\ln(\vert x\vert)+i\pi H(-x) $ and $$T_1=\frac{1}{x+i0}=\frac{d}{dx}(Log(x+i0))=pv\frac 1{x}-i\pi \delta_0(x).$$ It is easy to see that $ WF T_1=[0]\times (0,+\infty), $ so that $WF T_1+WF T_1$ does not meet 0. Then there is no difficulty to define $T^2$ say as $$ \langle T^2,\phi\rangle=\lim_{\epsilon\rightarrow 0_+}\int\frac{\phi(x) dx}{(x+i\epsilon)^2}. $$

(2) Let us consider a smooth hypersurface $\Sigma$ of $\mathbf R^d$ defined by the equation $f(x)=0$ with a smooth $f$ such that $df\not=0$ at $f=0$ and let $\delta_\Sigma$ be the Euclidean measure on $\Sigma$. Then $$ T_2=pv\frac{1}{f}-i\delta_\Sigma $$ can be squared. The reason is the same than for the previous example, since $ WF T_2$ is the positive conormal of $\Sigma$. A point $(x,\xi)\in WF T_2$ iff $$ x\in \Sigma\quad \xi =\lambda df(x) \text{ with $\lambda >0$}. $$ Then of course, if $(x,\xi_j)$, $j=1,2$ are both in $WF T_2$ then $$ \xi_1+\xi_2\not=0. $$

Denis Serre's answer is just perfect. Let me add a couple of examples of distributions that can be squared:

(1) With $H$ the Heaviside function, define $\operatorname{Log}(x+i0)=\ln(\vert x\vert)+i\pi H(-x) $ and $$T_1=\frac{1}{x+i0}=\frac{d}{dx}(\operatorname{Log}(x+i0)) = \operatorname{pv}\frac 1{x}-i\pi \delta_0(x).$$ It is easy to see that $ WF T_1=[0]\times (0,+\infty), $ so that $WF T_1+WF T_1$ does not meet $0.$ Then there is no difficulty to define $T^2$ say as $$ \langle T^2,\phi\rangle=\lim_{\varepsilon\rightarrow 0_+} \int\frac{\varphi(x) \, dx}{(x+i\varepsilon)^2}. $$

(2) Let us consider a smooth hypersurface $\Sigma$ of $\mathbf R^d$ defined by the equation $f(x)=0$ with a smooth $f$ such that $df\not=0$ at $f=0$ and let $\delta_\Sigma$ be the Euclidean measure on $\Sigma$. Then $$ T_2=pv\frac{1}{f}-i\delta_\Sigma $$ can be squared. The reason is the same than for the previous example, since $ WF T_2$ is the positive conormal of $\Sigma$. A point $(x,\xi)\in WF T_2$ iff $$ x\in \Sigma\quad \xi =\lambda df(x) \text{ with $\lambda >0$}. $$ Then of course, if $(x,\xi_j)$, $j=1,2$ are both in $WF T_2$ then $$ \xi_1+\xi_2\not=0. $$