Prove that $d(x+y, \mathcal{K}) \leq d(x, \mathcal{K})+d(y, \mathcal{K})$ for all $ x, y \in \mathcal{H}$

Question

Let $\mathcal{H}$ be a Hilbert space over $\mathbb{C},$ and let $\mathcal{K}$ be a closed subspace of $\mathcal{H}.$

Prove that $d(x+y, \mathcal{K}) \leq d(x, \mathcal{K})+d(y, \mathcal{K}) \forall x, y \in \mathcal{H}$

We have that there exists a unique vector in $\mathcal{K}$ that minimizes the distances between $x+y$ and $\mathcal{K},$ call that $a+b.$

There exists a sequence $a_n+b_n$ that satisfies the following, \begin{align} d(x+y,\mathcal{K}) &\leq \| (x+y) - (a_n+b_n) \| \\ &\leq \| x - a_n \| + \| y-b_n \| \\ &< d(x,\mathcal{K}) + \frac{1}{n} + d(y,\mathcal{K}) + \frac{1}{n}.\end{align} Then taking $n \rightarrow \infty, $ we have the desired inequality $d(x+y, \mathcal{K}) \leq d(x, \mathcal{K})+d(y, \mathcal{K}).$

Does this proof look correct, or close to correct? I could not find a similar question already posted.

Answer 1

Suppose $d_K(x) = \|x-a\|, d_K(y)= \|y-b\|$. Then $a+b \in K$ and so $d_K(x+y) \le \|x+y-(a+b)\| \le \|x-a\| + \|y-b\| = d_K(x)+d_K(y)$.

This is true whether or not $K$ is closed, you just need to refine the argument slightly.

It is straightforward to show that $d_K(tx) = |t| d_K(x)$ and since $K$ is closed, $d_K(x) = 0 $ iff $x \in K$.

Hence we can use $d_K$ to define a norm on $H / K$ (that is the quotient space where $x \sim y$ iff $x-y \in K$).

Answer 2

Let $\epsilon>0$. As $d(x,K)=\inf_{k\in K}d(x, k)$, there is a unique $a\in K$ such that $$\|x - a\|= d(x, a) < d(x, K) + \epsilon$$. Similarly, there is a unique $b\in K$ such that $$\|y - b\| = d(y, b) < d(y, K) + \epsilon.$$ Since $K$ is a subspace of $H$, $a + b\in K$. Therefore, $$ \begin{aligned} d(x+y, K)&\leq d(x+y, k_x + k_y)\\ &= \|(x+y) - (k_x+k_y)\|\\ &= \|(x - k_x) + (y - k_y)\|\\ &\leq \|x - a\| + \|y - b\|\\ &= d(x, a) + d(y, b)\\ &< d(x, K) + d(y, K) + 2\epsilon. \end{aligned} $$ Since this is true for all $\epsilon > 0$, we must have $$ d(x+y, K)\leq d(x, K) + d(y, K). $$