left and right support If you consider a bounded linear operator $x$, why is it true that the smallest projection $p$ such that $px=x$ is $uu^* $ where $u$ is given by the polar decomposition of $x$? Moreover why the smallest projection $q$ s.t. $xq=x$ is $u^*u$. 
It follows from the construction that $ker(u)=ker(v)$ and so that $u^*u $ is the projection onto the closure of $u^*(H)$ which I thought may be usefull but it doesn't seem to help unless you know what the partial isometry coming from the polar decomposition of $v^*$ is.
 A: Let $x \in \mathcal L(\mathcal H)$ be a bounded, linear operator. Then, $x = u|x|$ is the (unique) polar decomposition of $x$ with $|x| := (x^*x)^{1/2}$ and $u$ a partial isometry which means that $\left.u\right|_{\operatorname{ker} u^\perp}$ is isometric. Set $\mathcal M = {\operatorname{ker} u^\perp}$ and $\mathcal N = \overline{\operatorname{ran}(u)}$. Then $u$ is identically $0$ on $\mathcal M^\perp$ and $u$ maps $\mathcal M$ isometrically onto $\mathcal N$. 
First of all, we show that $p_1 = u^*u$ is a projection. For $i=1,2$ consider $a_i \in \mathcal H$ and $b_i \in \mathcal M$, $c_i \in \mathcal M^\perp$ with $a_i = b_i + c_i$. Then $\langle u^*u a_1, a_2 \rangle = \langle u a_1, u a_2 \rangle = \langle u b_1, u b_2 \rangle = \langle  b_1,  b_2 \rangle = \langle  b_1,  a_2 \rangle$.
Next, note that also $p_2 = uu^*$ becomes a projection since $(u^*u)² = u^*u \iff uu^*u= u \iff u^*uu^* = u^* \iff (uu^*)^2 = uu^*$. Now it's easy to see that $u^*$ is also a partial isometry with $\operatorname{ker}(u^*) = \mathcal N^\perp$ and $\overline{\operatorname{ran}(u^*)} = \mathcal M$.
Define $\mathit l(x) = \inf\{p: p \text{ a projection}, px = x\}$ and $\mathit r(x) = \inf\{p: p \text{ a projection}, xp = x\}$. Observe that $\mathit l(x) = \text{projection onto } \overline{\operatorname{ran}(x)}$ and $\mathit r(x) = \mathit{l}(x^*)$.
Summarizing, $uu^*$ is a projection onto $\mathcal N$ which is by definition $\overline{\operatorname{ran}(x)}$, i.e. $\mathit{l}(x) = uu^*$. Since $x^* = u^*|x^*|$, $\mathit{l}(x^*) = u^*u = \mathit{r}(x)$.
