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By Hilbert projection theorem, if $x\in\mathbb{R}^n$ and $D$ is a closed subspace of $\mathbb{R}^n$ then the optimization problem $$\underset{y}{\min} \|x-y\| \ s.t. \ y \in D \quad\quad (P1)$$ has an unique solution, namely, the $\bar{x} \in \mathbb{R}^n$ such that $x-\bar{x}$ is orthogonal to $D$.

Consider now that $D$ is a compact differentiable manifold. Since $f(y):=\|x-y\|$ is a continuous function on the compact $D$, the optimization problem (P1) has a global minimum $\bar{x}$.

It must be the case that $x-\bar{x}$ is normal to the tangent space of $D$ at $\bar{x}$?

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If $x\notin D$ then yes. To see this consider smooth curves $\gamma(t)\subset D$ passing through $\bar x$ in $t=0$ and look at $$h(t)= f(\gamma(t))^2$$ This function clearly attains it's minimum at $t=0$, so it's derivative is zero at $t=0$.

It's rather easy to see that this derivative is given by $$h^\prime(0) = 2\langle x-\bar x, \gamma^\prime\rangle = 0$$ and since this is true for each such $\gamma$ the scalar product vanishes for each tangent vector to $D$ in $\bar x$.

(Looking at the squared norm just simplifies the computation of the derivative, you could work with $f$, too).

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The statement of Hilbert projection theorem in your question is not correct. The set $D$ need to be a closed "subspace" of the Hilbert space. In the case of $\mathbb{R}^n$, a line, plan,...etc are examples of subspaces. However, a unit circle for example on $\mathbb{R}^2$ is a compact differential manifold but not a subspace.

For the existence of the minimizer, you need at least the set $D$ to be closed, non-empty and convex.

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    $\begingroup$ You are right. Thanks, I've edited the question accordingly to your comment. $\endgroup$
    – shamisen
    Commented Mar 24, 2018 at 3:36
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    $\begingroup$ This link might answer your question: mathoverflow.net/questions/242002/… $\endgroup$
    – bersou
    Commented Mar 24, 2018 at 22:52
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    $\begingroup$ Bersou thanks again for the help :) $\endgroup$
    – shamisen
    Commented Mar 26, 2018 at 0:59

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