[proofplan]
We prove existence by constructing a minimising sequence for $\|v - w\|_V$ over $w \in K$, then use the parallelogram law together with convexity of $K$ to show this sequence is Cauchy. Completeness of $V$ and closedness of $K$ give convergence to a minimiser $u \in K$. We prove uniqueness by applying the same parallelogram-law argument to two hypothetical minimisers, showing their distance is zero.
[/proofplan]
[step:Construct a minimising sequence for the distance to $v$]
Let $d := \inf_{w \in K} \|v - w\|_V$. By the definition of infimum, there exists a sequence $(w_k)_{k=1}^\infty \subset K$ with
\begin{align*}
\lim_{k \to \infty} \|v - w_k\|_V = d.
\end{align*}
[/step]
[step:Apply the parallelogram law to show $(w_k)$ is Cauchy]
The parallelogram law in the [Hilbert space](/page/Hilbert%20Space) $V$ states that for any $a, b \in V$:
\begin{align*}
\|a + b\|_V^2 + \|a - b\|_V^2 = 2\|a\|_V^2 + 2\|b\|_V^2.
\end{align*}
Apply this with $a = v - w_k$ and $b = v - w_j$:
\begin{align*}
\|(v - w_k) + (v - w_j)\|_V^2 + \|w_j - w_k\|_V^2 = 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2.
\end{align*}
The first term on the left satisfies
\begin{align*}
\|(v - w_k) + (v - w_j)\|_V^2 = 4\left\|v - \frac{w_k + w_j}{2}\right\|_V^2.
\end{align*}
Since $K$ is convex, $\frac{w_k + w_j}{2} \in K$, so by definition of $d$:
\begin{align*}
\left\|v - \frac{w_k + w_j}{2}\right\|_V \ge d.
\end{align*}
Substituting and rearranging:
\begin{align*}
\|w_k - w_j\|_V^2 &= 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2 - 4\left\|v - \frac{w_k + w_j}{2}\right\|_V^2 \\
&\le 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2 - 4d^2.
\end{align*}
As $k, j \to \infty$, both $\|v - w_k\|_V^2$ and $\|v - w_j\|_V^2$ converge to $d^2$, so
\begin{align*}
\limsup_{k, j \to \infty} \|w_k - w_j\|_V^2 \le 2d^2 + 2d^2 - 4d^2 = 0.
\end{align*}
Therefore $(w_k)$ is a [Cauchy sequence](/page/Cauchy%20Sequence).
[guided]
The parallelogram law is the essential tool here — it is the identity that distinguishes [Hilbert spaces](/page/Hilbert%20Space) from general [Banach spaces](/page/Banach%20Space), and it is what makes the projection unique.
We apply the parallelogram law with $a = v - w_k$ and $b = v - w_j$:
\begin{align*}
\|(v - w_k) + (v - w_j)\|_V^2 + \|(v - w_k) - (v - w_j)\|_V^2 = 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2.
\end{align*}
The second term on the left simplifies to $\|w_j - w_k\|_V^2$, which is what we want to bound. The first term:
\begin{align*}
\|(v - w_k) + (v - w_j)\|_V^2 = \left\|2\left(v - \frac{w_k + w_j}{2}\right)\right\|_V^2 = 4\left\|v - \frac{w_k + w_j}{2}\right\|_V^2.
\end{align*}
Now convexity of $K$ enters: since $w_k, w_j \in K$ and $K$ is convex, the midpoint $\frac{w_k + w_j}{2} \in K$. By definition of $d = \inf_{w \in K}\|v - w\|_V$, we get $\|v - \frac{w_k + w_j}{2}\|_V \ge d$. Rearranging:
\begin{align*}
\|w_k - w_j\|_V^2 &= 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2 - 4\left\|v - \frac{w_k + w_j}{2}\right\|_V^2 \\
&\le 2\|v - w_k\|_V^2 + 2\|v - w_j\|_V^2 - 4d^2.
\end{align*}
Since $\|v - w_k\|_V \to d$ and $\|v - w_j\|_V \to d$, the right-hand side tends to $2d^2 + 2d^2 - 4d^2 = 0$. Therefore $\|w_k - w_j\|_V \to 0$, and $(w_k)$ is Cauchy.
What would fail in a general [Banach space](/page/Banach%20Space)? The parallelogram law does not hold, so we cannot bound $\|w_k - w_j\|_V^2$ below from the midpoint distance. Indeed, the projection onto a closed convex set need not be unique in a general Banach space.
[/guided]
[/step]
[step:Conclude existence by completeness and closedness]
Since $V$ is a [Hilbert space](/page/Hilbert%20Space), it is [complete](/page/Complete%20Metric%20Space). The [Cauchy sequence](/page/Cauchy%20Sequence) $(w_k)$ converges to some $u \in V$. Since $K$ is closed and $(w_k) \subset K$, the limit $u$ belongs to $K$. By continuity of the norm:
\begin{align*}
\|v - u\|_V = \lim_{k \to \infty} \|v - w_k\|_V = d = \inf_{w \in K} \|v - w\|_V.
\end{align*}
Thus $u \in K$ achieves the infimum.
[/step]
[step:Prove uniqueness via the parallelogram law]
Suppose $u_1, u_2 \in K$ both satisfy $\|v - u_1\|_V = \|v - u_2\|_V = d$. Since $K$ is convex, $\frac{u_1 + u_2}{2} \in K$, so $\|v - \frac{u_1 + u_2}{2}\|_V \ge d$.
Apply the parallelogram law with $a = v - u_1$ and $b = v - u_2$:
\begin{align*}
4\left\|v - \frac{u_1 + u_2}{2}\right\|_V^2 + \|u_1 - u_2\|_V^2 = 2\|v - u_1\|_V^2 + 2\|v - u_2\|_V^2 = 4d^2.
\end{align*}
Rearranging:
\begin{align*}
\|u_1 - u_2\|_V^2 = 4d^2 - 4\left\|v - \frac{u_1 + u_2}{2}\right\|_V^2 \le 4d^2 - 4d^2 = 0.
\end{align*}
Therefore $\|u_1 - u_2\|_V = 0$, so $u_1 = u_2$.
[/step]
