\begin{proof} We first show 1. Let $x\in H$, $m_{0}\in B$. Want to show that if $x-m_{0}$ is not Orthogonal vector to $B$ then it cannot be optimal.

Suppose $m_{0}\in B$ is such that $\exists m\in B$ with $\left< x-m_{0},m \right>>0$, i.e., $x-m_{0}$ is not orthogonal to $B$.

WLOG, take $\lVert m \rVert=1$ and $\left< x-m_{0},m \right>=\delta$.

We can show that with $m_{1}=m_{0}+\delta m$, we will have $$\begin{align*} \lVert x-m_{1} \rVert ^{2}&= \sqrt{ \langle x-m_{1}, x-m_{1} \rangle }^{2}\\ &= \langle x-m_{0}-\delta m, x-m_{0}-\delta m \rangle \\ &= \langle x-m_{0}, x-m_{0} \rangle -\delta \langle m, x-m_{0}\rangle-\delta \langle x-m_{0}, m \rangle +\delta^{2}\langle m, m \rangle \\ &= \lVert x-m_{0} \rVert ^{2}-2\delta^{2}+\delta^{2}\\ &= \lVert x-m_{0} \rVert ^{2}-\delta^{2}\\ &< \lVert x-m_{0} \rVert ^{2} \end{align*}$$and thus $m_{0}$ cannot be a minimizer of $(1)$.

We now show that if $x-m^{*}$ is orthogonal to $B$, then it is optimal. If $\left< x-m^{*},M \right> =0$, we have that for any $m\in B$
$$\lVert x-m \rVert ^{2}=\lVert x-m^{*}+m^{*}-m \rVert ^{2}=\lVert x-m^{*} \rVert ^{2}+\lVert m^{*}-m \rVert ^{2}\ge \lVert x-m^{*} \rVert^{2}$$ since for $x\perp y$ we have that $\lVert x+y \rVert^{2}=\lVert x \rVert^{2}+\lVert y \rVert^{2}$. This yields a contradiction, proving 1.

We now show 2. Let $$\delta=\inf_{m\in B}\lVert x-m \rVert$$Let $\{ m_{k} \}_{k\in \mathbb{N}}$ be s.t. $$\lVert x-m_{k} \rVert \to\delta$$and observe that $$\begin{align*} &\langle x-m_{k}+x-m_{n}, x-m_{k}+x-m_{n} \rangle +\langle x-m_{k}-(x-m_{n}), x-m_{k}-(x-m_{n}) \rangle \\ &= \langle x-m_{k}, x-m_{k} +x-m_{n}\rangle +\langle x-m_{n}, x-m_{k}+x-m_{n} \rangle \\ &\quad\quad\quad+ \langle x-m_{k}, x-m_{k}-(x-m_{n}) \rangle +\langle -(x-m_{n}), x-m_{k}-(x-m_{n}) \rangle \\ &= 2\langle x-m_{k}, x-m_{k} \rangle +2\langle x-m_{n}, x-m_{n} \rangle \end{align*}$$Write $$\begin{align*} \langle x-m_{k}+x-m_{n}, x-m_{k}+x-m_{n} \rangle &= \left\langle 2\left( x- \frac{m_{k}+m_{n}}{2} \right), 2\left( x- \frac{m_{k}+m_{n}}{2} \right) \right\rangle \\ &= 4\left\langle x- \frac{m_{k}+m_{n}}{2}, x- \frac{m_{k}+m_{n}}{2} \right\rangle \tag{🚧} \end{align*}$$Since $\frac{m_{k}+m_{n}}{2}\in B$ then by our definition of $\delta$ we have that: $$\left\lVert x- \frac{m_{k}+m_{n}}{2} \right\rVert \ge\delta$$then we square both sides then multiply by $4$ $$4\left\lVert x- \frac{m_{k}+m_{n}}{2} \right\rVert ^{2}\ge 4\delta^{2}$$ and since the LHS is exactly $(🚧)$ then by applying Parallelogram law we get $$\lVert m_{k}-m_{n} \rVert ^{2}\le 2\lVert x-m_{n} \rVert ^{2}+2\lVert x-m_{k} \rVert ^{2}-4\delta^{2}$$As a result, as $\lVert x-m_{n} \rVert^{2}\to\delta^{2}$, we have that $m_{k}$ is Cauchy. Since $B$ is closed, it has a limit, call the limit $\tilde{m}$. We claim that the limit is optimal and hence $\tilde{m}=m^{*}$. Conside the difference: $$\langle x-m_{n}, x-m_{n} \rangle -\langle x-\tilde{m}, x-\tilde{m} \rangle$$ We claim that the difference goes to zero; this follows from the Continuity of inner product (Yuksel). Indeed $$\begin{align*} \left| \langle x-m_{n}, x-m_{n} \rangle -\langle x-\tilde{m}, x-\tilde{m} \rangle \right| &= \left| \langle x-m_{n}, x-m_{n} \rangle -\langle x-m_{n}, x-\tilde{m} \rangle\right.\\ &\quad\quad\quad\left.+ \langle x-m_{n}, x-\tilde{m} \rangle- \langle x-\tilde{m}, x-\tilde{m} \rangle \right| \\ &\le \left| \langle x-m_{n}, x-m_{n} \rangle -\langle x-m_{n}, x-\tilde{m} \rangle \right| \\ &\quad\quad+ \left| \langle x-m_{n}, x-\tilde{m} \rangle -\langle x-\tilde{m}, x-\tilde{m} \rangle \right| \\ &= \left| \langle x-m_{n}, m_{n}-\tilde{m} \rangle \right| +\left| \langle m_{n}-\tilde{m}, x-\tilde{m} \rangle \right| \\ &\le \lVert x-m_{n} \rVert \lVert m_{n} -\tilde{m}\rVert +\lVert m_{n}-\tilde{m} \rVert \lVert x-\tilde{m} \rVert \end{align*}$$where the final inequality is due to Cauchy-Schwartz. Now, since $\lVert x-m_{n} \rVert\to\delta$, we have that $\lVert x-m_{n} \rVert$ is bounded. Finally as $\lVert m_{n}-\tilde{m} \rVert\to0$, both terms in the final line go to zero and we conclude that $$\delta=\lim_{ n \to \infty } \lVert x-m_{n} \rVert =\lVert x-\tilde{m} \rVert$$and therefore $\tilde{m}$ is a minimizing vector. By part 1, this has to be the only minimizing vector in $B$.

\end{proof}

\begin{proof} Let $D=\{ x_{1},x_{2},\dots ,\}$ be a countable set dense in $H$. Let $\{ e_{\alpha} \}_{\alpha \in \mathcal{A}}\subset H$ be a set of orthonormal vectors. Observe that by , for $\alpha \neq\beta$:$$\lVert e_{\alpha}-e_{\beta} \rVert^{2}=\lVert e_{\alpha} \rVert^{2}+\lVert e_{\beta} \rVert^{2} =2$$hence $$\lVert e_{\alpha}-e_{\beta} \rVert =\sqrt{ 2 }$$ By denseness of $D$, for every $e_{\alpha}$, there exists some element $x_{k_{\alpha}}\in D$, such that $$\lVert e_{\alpha}-x_{k_{\alpha}} \rVert < \frac{1}{\sqrt{ 2 }}.$$this is because we want to select $x_{k_{\alpha}}$ such that it is injective wrt $e_{\alpha}$ and hence we need to restrict the radius using the following relation $$2\epsilon< \sqrt{ 2 }\iff\epsilon< \frac{1}{\sqrt{ 2 }}.$$ Let $M$ be the collection of such vectors $\{ x_{k_{\alpha}} \}_{\alpha \in \mathcal{A}}\subset D$ and observe that since it is a subset of $D$, this set is countable as well. Now, for a given $x_{k_{\alpha}}$, for any $e_{\beta}$ with $\beta \neq\alpha$, we have, by the relation, $\lVert x \rVert-\lVert y \rVert\le \lVert x-y \rVert$ $$\lVert e_{\beta}-e_{\alpha} \rVert -\lVert -(e_{\alpha}-x_{k_{\alpha}}) \rVert \le \lVert e_{\beta}-e_{\alpha}-(-(e_{\alpha}-x_{k_{\alpha}})) \rVert =\lVert e_{\beta}-x_{k_{\alpha}} \rVert$$and thus $$\sqrt{ 2 }- \frac{1}{\sqrt{ 2 }}= \frac{1}{\sqrt{ 2 }}\le \lVert e_{\beta}-x_{k_{\alpha}} \rVert$$Therefore, for every $x_{k_{\alpha}}\in M, !\exists e_{\alpha}:d(x_{k_{\alpha}},e_{\alpha})< \frac{1}{\sqrt{ 2 }}$. Thus, we can associate with every element in $\{ e_{\alpha} \}_{\alpha \in \mathcal{A}}$ a unique element in the countable set $M\subset D$. Thus, the set $\{ e_{\alpha} \}_{\alpha \in \mathcal{A}}$ is countable. \end{proof} >[!def|2.3.5] Complete orthonormal sequence >An orthonormal sequence in a Hilbert space $H$ is complete if the only vector in $H$ which is orthogonal to each of the vectors is the null vector.

\begin{proof} For $n\in \mathbb{N}$ let $$x_{n}=\sum_{i=1}^{n}\langle x, e_{i} \rangle e_{i}.$$This vector $x_{n}$ is the projection of $x$ onto the subspace spanned by the first $n$ basis vectors. The vector $x-x_{n}$ is orthogonal to this subspace, and thus orthogonal to $x_{n}$ itself. By the we have $$\lVert x \rVert ^{2}=\lVert x_{n} \rVert ^{2}+\lVert x-x_{n} \rVert ^{2}.$$The proof proceeds by computing $\lVert x-x_{n} \rVert^{2}$: $$\begin{align*} 0&\le \left\lVert x-\sum_{i=1}^{n}\langle x, e_{i} \rangle e_{i} \right\rVert ^{2}\\ &= \left\langle x-\sum_{i=1}^{n}\langle x, e_{i} \rangle e_{i}, x-\sum_{j=1}^{n}\langle x, e_{j} \rangle e_{j} \right\rangle \\ &= \lVert x \rVert ^{2}-\sum_{i=1}^{n} \overline{\langle x, e_{i} \rangle }\langle x, e_{i} \rangle -\sum_{j=1}^{n}\langle x, e_{j} \rangle \langle e_{j}, x \rangle +\sum_{i=1}^{n}\sum_{j=1}^{n}\overline{\langle x, e_{i} \rangle }\langle x, e_{j} \rangle \langle e_{i}, e_{j} \rangle \\ &= \lVert x \rVert ^{2}-\sum_{i=1}^{n}\left| \langle x, e_{i} \rangle \right| ^{2}-\sum_{j=1}^{n}\left| \langle x, e_{j} \rangle \right| ^{2}+\sum_{i=1}^{n}\left| \langle x, e_{i} \rangle \right| ^{2}\\ &= \lVert x \rVert ^{2}-\sum_{i=1}^{n}\left| \langle x, e_{i} \rangle \right| ^{2} .\end{align*}$$This holds for any $n\ge 1$. Since the partial sums are non-negative and bounded above by $\lVert x \rVert^{2}$, the series $\sum_{i=1}^{\infty}\left| \langle x, e_{i} \rangle \right|^{2}$ converges and its sum is less than or equal to $\lVert x \rVert^{2}$. \end{proof}

\begin{proof} $\impliedby$: Let $H$ be separable. Then, there exists a countable dense subset $D=\{ x_{1},x_{2},\dots \}$. Apply the Gran-Schmidt process (212) to obtain $\{ e_{1},e_{2},\dots, \}$, an orthonormal basis. We claim that this set is a complete orthonormal sequence.

Suppose not; that is, let $h\in H$ be so that $\lVert h \rVert\neq 0$ and let $$\langle h, e_{k} \rangle=0,\forall k\in \mathbb{N}\tag{⭐️}.$$i.e., let $h$ be orthogonal to each $e_{i}$ and not the null vector. Now, for every $\epsilon>0$, there exists $x_{m}\in D$ with $$\lVert h-x_{m} \rVert\le \epsilon\tag{3}$$ and observe that $$x_{m}=\sum_{k=1}^{m}\alpha_{i}e_{i}\tag{4}$$since the span of vectors $\{ e_{1},\dots,e_{m} \}$ contains $x_{m}$. Then, $$\lVert h \rVert ^{2}=\langle h, h \rangle =\left\langle h-\sum_{k=1}^{m}\alpha_{i}e_{i}, h \right\rangle =\langle h-x_{m}, h \rangle \le \lVert h-x_{m} \rVert \lVert h \rVert \le \epsilon \lVert h \rVert$$ which is pretty much: Inner product defines a norm (Yuksel), add subtract summation term and apply linearity of inner product then apply $(⭐️)$, $(4)$, Cauchy-Schwarz Inequality (Yuksel), then $(3)$. This implies that $\lVert h \rVert\le \epsilon$. Since $\epsilon>0$ is arbitrary; then we have $\lVert h \rVert=0$, a contradiction hence $\lVert h \rVert=0$ and $h$ is the null element.

$\implies$ Now, let $H$ have a complete orthonormal sequence $\{ e_{1},e_{2},\dots \}$. We will show that $$D=\bigcup_{n\in \mathbb{N}}\left\{ x\in H: x=\sum_{i=1}^{n}\alpha_{i}e_{i},\alpha_{i}\in \mathbb{Q} \right\},$$is a

1. countable
2. dense subset in $H$, and separability of $H$ follows by the definition.

1. That $D$ is countable follows from the fact that for every $n$, the set $\left\{ x\in H: x=\sum_{i=1}^{n}\alpha_{i}e_{i},\alpha_{i}\in \mathbb{Q} \right\}$ is countable as a Cartesian product of finitely many countable sets, and thus the countable union over $n\in \mathbb{N}$ leads to a countable set.

2. We now show that this set is dense in $H$. i.e., that $\forall x\in H, \forall\epsilon>0, \exists e\in D:\lVert x-e\rVert<\epsilon$ where $e=\sum_{i=1}^{M}\alpha_{i}e_{i},\alpha_{i}\in \mathbb{Q}$.

Consider the sequence of vectors $x_{n}:=\sum_{i=1}^{n}\langle x, e_{i} \rangle e_{i}$. Observe that $x_{n}$ converges to a limit by Theorem 2.3.1. To see that the sequence $\sum_{i=1}^{\infty}\left| \langle x, e_{i} \rangle \right|^{2}$ is summable, use to conclude $$\sum_{i=1}^{\infty}\left| \langle x, e_{i} \rangle \right| ^{2}\le \lVert x \rVert ^{2}.$$Therefore, the limit $\sum_{i=1}^{\infty}\langle x, e_{i} \rangle e_{i}$ is well-defined and hence $x_{n}\to x$ for some $x\in H$.

Now, let $h=x-\sum_{i=1}^{\infty} \langle x, e_{i} \rangle e_{i}$. we claim that $\lVert h \rVert=0$. For any $m\in \mathbb{N}$, $$\begin{align*}\langle h, e_{m} \rangle &= \left\langle x-\sum_{i=1}^{\infty}\langle x, e_{i} \rangle e_{i}, e_{m} \right\rangle \\ &= \langle x, e_{m} \rangle -\left\langle \lim_{ n \to \infty } \sum_{i=1}^{n}\langle x, e_{i} \rangle e_{i}, e_{m} \right\rangle \\ &= \langle x, e_{m} \rangle -\lim_{ n \to \infty } \left\langle \sum_{i=1}^{n}\langle x, e_{i} \rangle e_{i}, e_{m} \right\rangle \\ &= \langle x, e_{m} \rangle -\langle x, e_{m} \rangle \\ &= 0 \end{align*}$$ But since $\{ e_{1},e_{2},\dots \}$ is a complete orthonormal sequence, it must be that $\lVert h \rVert=0$. Hence, $x=\sum_{i=1}^{\infty}\langle x, e_{i} \rangle e_{i}$. Now approximate this vector, by first truncating the sum so that $\forall\epsilon>0$, $$\left\lVert x-\sum_{i=1}^{N}\langle x, e_{i} \rangle e_{i} \right\rVert \le \frac{\epsilon}{2}$$and then approximating the coefficients by rational numbers so that $$\left\lVert \sum_{i=1}^{N}\langle x, e_{i} \rangle e_{i}-\sum_{i=1}^{N}\alpha_{i}e_{i} \right\rVert \le \frac{\epsilon}{2},\quad\alpha_{i}\in \mathbb{Q}.$$This implies $$\begin{align*} \left\lVert x-\sum_{i=1}^{N}\alpha_{i}e_{i} \right\rVert &\le \left\lVert x-\sum_{i=1}^{N}\langle x, e_{i} \rangle e_{i} \right\rVert +\left\lVert \sum_{i=1}^{N}\langle x, e_{i} \rangle e_{i}-\sum_{i=1}^{N}\alpha_{i}e_{i} \right\rVert \\ &\le \epsilon \end{align*}$$Since for every $\epsilon$ such an approximation can be made by some element in $D$, this completes the proof.

\end{proof}