The L2 projection onto a pointwise box is clipping

Put $q(x)=\min \{q_b(x),\max \{q_a(x),r(x)\}\}$. This function is measurable and lies between $q_a$ and $q_b$; therefore $q\in Q_{ad}$ and $q\in L^2(E)$. For real numbers $a\le b$, the clipped value $c=\min \{b,\max \{a,s\}\}$ satisfies $(c-s)(t-c)\ge 0$ for every $t\in [a,b]$: if $s<a$, then $c=a$ and both factors are nonnegative; if $a\le s\le b$, then $c=s$ and the product is zero; if $s>b$, then $c=b$ and both factors are nonpositive. Applying this pointwise with $s=r(x)$ and $t=v(x)$ and integrating gives $(q-r,v-q{)}_{L^2}\ge 0$ for every $v\in Q_{ad}$.

For any $v\in Q_{ad}$, $\mathrm{\parallel }v-r{\mathrm{\parallel }}_2^2-\mathrm{\parallel }q-r{\mathrm{\parallel }}_2^2=\mathrm{\parallel }v-q{\mathrm{\parallel }}_2^2+2(q-r,v-q{)}_2\ge \mathrm{\parallel }v-q{\mathrm{\parallel }}_2^2\mathrm{.}$ Thus $q$ minimizes the distance and equality can hold only when $v=q$ in $L^2$. Conversely, if $q$ is a distance minimizer, then for any $v\in Q_{ad}$ and $0<t\le 1$, $q+t(v-q)\in Q_{ad}$. Expanding $\mathrm{\parallel }q+t(v-q)-r{\mathrm{\parallel }}_2^2\ge \mathrm{\parallel }q-r{\mathrm{\parallel }}_2^2$, dividing by $2t$, and letting $t\downarrow 0$ gives $(q-r,v-q{)}_2\ge 0$.