I am a student of MSc second year,Mathematics.I always try to produce visually intuitive proofs so that it is easy to remember.Recently,we were taught a theorem in functional analysis which is as follows:
Statement: Let $H$ be a Hilbert space and $M$ be a closed subspace then $H=M\oplus M^{\perp}$.
My attempt: Let $x\in H$ ,we want to show $x\in M+M^{\perp}$.Let $y\in M$ be the unique vector such that $\|x-y\|\leq \|x-y'\|$ for all $y'\in M$.Then write $x=(x-y)+y$.If we could show that $x-y\in M^{\perp}$,we are through.So let $z\in M$ ,we have to show $\langle x-y,z\rangle=0$.On the contrary,let us assume that $\langle x-y,z\rangle\neq 0$ which pictorially means that $x-y$ is not perpendicular to $z$,so it will have a non-zero projection along $z$,which is given by $\frac{\langle x-y,z\rangle}{\|z\|^2}z$.Then it is intuitive that $x-y-\frac{\langle x-y,z\rangle}{\|z\|^2}z$ will be perpendicular to $z$,which can also be checked formally.But now,I am not able to figure out how to proceed.Can someone give me some hint?My guess is that this vector $x-y-\frac{\langle x-y,z\rangle}{\|z\|^2}z\in M^{\perp}$ but I am not able to show that.Even if I do,I have no way to proceed further.
Addendum
As mentioned by Chee Han.I provide here some background of what has been done before this theorem.Basically a lemma has been proved which says:
If $M$ is a closed subspace of a Hilbert space $H$,then for each $x\in H$ there exists a unique $y\in M$ such that $\|x-y\|\leq \|x-y'\|$ for all $y'\in M$.
Using this,I have to prove this theorem.
Given that you have the minimization property, this is easy enough (the tough part, which uses completeness, is the lemma which tells you the existence of a unique $y$).
Ok, so let $x\in H$ be given, and let $y\in M$ be the unique element given by your lemma. You’re right, we now write $x=y+(x-y)$, and we wish to show the second part is in $M^{\perp}$. To do this, consider any $z\in M$, and consider the function $f:\Bbb{R}\to\Bbb{R}$ defined as $f(t)=\|x-y+tz\|^2$. Then, $f$ is a quadratic polynomial in $t$, so it is differentiable everywhere, and furthermore, $y,z\in M$ implies $y-tz\in M$; so by the minimizing property of $y$, it follows that $f$ has a (global) minimum at $t=0$. You can now differentiate at $t=0$ to get $\langle x-y,z\rangle=0$. This shows $x-y\in M^{\perp}$.
The idea is that a minimization problem is of course closely related to derivatives. Here, $M$ was a subspace which was crucial for us to say $y-tz\in M$ for all $t\in\Bbb{R}$, and hence $f(0)=\|x-y\|^2\leq \|x-(y-tz)\|^2=f(t)$, and thus $f’(0)=0$.
As a bonus: note that the existence of a unique minimizing $y$ also holds if you assume only that $M$ is a closed convex subset (when $M$ is a subspace, convexity is automatic). Try to derive the analogous condition on inner products in this case (hint: an inequality will be present, since convexity means that (a slight modification of) the above function is only defined for $t\in [0,1]$ and we have a minimum at an endpoint of the interval).