Closed l two subspaces have orthogonal projections

Since $0\in M$, $d={\inf }_{g\in M}\mathrm{\parallel }f-g\mathrm{\parallel }$ exists in $[0,\mathrm{\parallel }f\mathrm{\parallel }]$. For every $n\in \mathbb{N}$ the set $\{g\in M:\mathrm{\parallel }f-g{\mathrm{\parallel }}^2<d^2+1\mathrm{/}(n+1)\}$ is nonempty by the defining property of the infimum. AC selects $g_n$ in these sets, including $g_0$. It also supplies the countable choice assumed in complex completeness.

Expanding the pairing gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}^2+\mathrm{\parallel }u-v{\mathrm{\parallel }}^2=2\mathrm{\parallel }u{\mathrm{\parallel }}^2+2\mathrm{\parallel }v{\mathrm{\parallel }}^2$: the two cross terms cancel. Apply this to $u=f-g_n$, $v=f-g_m$. Since $(g_n+g_m)\mathrm{/}2\in M$, it follows that $\mathrm{\parallel }{g}_n-g_m{\mathrm{\parallel }}^2\le 2\mathrm{\parallel }f-g_n{\mathrm{\parallel }}^2+2\mathrm{\parallel }f-g_m{\mathrm{\parallel }}^2-4d^2<2\mathrm{/}(n+1)+2\mathrm{/}(m+1)$. Thus $(g_n{)}_{n\in \mathbb{N}}$ is Cauchy.

Completeness gives a norm limit in $L^2$. Closedness of $H$, then of $M$ in $H$, puts this limit $g$ in $M$. The triangle inequality implies $\mathrm{\mid }\mathrm{\parallel }f-g_n\mathrm{\parallel }-\mathrm{\parallel }f-g\mathrm{\parallel }\mathrm{\mid }\le \mathrm{\parallel }{g}_n-g\mathrm{\parallel }$, so $\mathrm{\parallel }f-g\mathrm{\parallel }=d$. This reasoning applies equally when $H$ is the full space.

Set $z=f-g$. For $h\in M$ and real $t$, minimality gives $0\le \mathrm{\parallel }z-th{\mathrm{\parallel }}^2-\mathrm{\parallel }z{\mathrm{\parallel }}^2=-2t\mathrm{Re}⟨z,h⟩+t^2\mathrm{\parallel }h{\mathrm{\parallel }}^2$. Dividing separately for positive and negative $t$ and letting $t\to 0$ makes the real part zero. Replacing $h$ by $ih\in M$ makes the imaginary part zero because $⟨z,ih⟩=-i⟨z,h⟩$. Hence $z\perp M$.

If $g^{\mathrm{\prime }}$ also lies in $M$ with $f-g^{\mathrm{\prime }}\perp M$, then $g-g^{\mathrm{\prime }}\in M\cap M^{\perp }$; its squared norm is zero, so $g=g^{\mathrm{\prime }}$. Define $P_{M}f=g$. Conversely any such orthogonal decomposition minimizes distance: for $h\in M$, expansion yields $\mathrm{\parallel }f-h{\mathrm{\parallel }}^2=\mathrm{\parallel }f-g{\mathrm{\parallel }}^2+\mathrm{\parallel }g-h{\mathrm{\parallel }}^2\ge \mathrm{\parallel }f-g{\mathrm{\parallel }}^2$. Thus it has exactly the required distance property.

For $a,b\in \mathbb{C}$ and $f_1,f_2\in H$, the vector $a{P}_M{f}_1+b{P}_M{f}_2$ lies in $M$, and its difference from $a{f}_1+b{f}_2$ is orthogonal to $M$ by sesquilinearity. Uniqueness gives linearity. Orthogonal expansion gives $\mathrm{\parallel }f{\mathrm{\parallel }}^2=\mathrm{\parallel }{P}_{M}f{\mathrm{\parallel }}^2+\mathrm{\parallel }f-P_{M}f{\mathrm{\parallel }}^2$, proving contraction. Pairing with each fixed $h\in M$ is continuous by Cauchy–Schwarz, so $M^{\perp }$ is closed; it is a subspace by linearity. Each $f$ has the displayed decomposition, and its uniqueness follows from $M\cap M^{\perp }=\{0\}$. When $M=\{0\}$ this gives $P_M=0$, and when $M=H$ it gives $P_M=I$, including the zero-space case.