show that $x\longmapsto \arg \min\|x-y\| $ is continuous

Question

Let $\varnothing\ne M\subset\mathbb R^n$ be a convex compact set, suppose $M\subseteq B_r(u)$ (closed ball).

Then the map \begin{align} f:\ B_r(u)&\longrightarrow M \newline x&\longmapsto f(x) = \arg\min_{y\in M}\|x-y\| \end{align} is continuous.

I was given a proof as followed :

First, the map $f$ is well-difined due to $f(x)$ exists and is unique.

For any $x\in B_r(u)$ and any sequence $x_n\to x$, by definition of $f$ we have \begin{align} \|x_n-f(x_n)\|\leq \|x_n-y\|, \ \forall y\in M;\tag1 \end{align} Hence, if $a$ is any limit of $f(x_n)$ then \begin{align} \|x-a\|\leq \|x-y\|, \ \forall y\in M. \tag2 \end{align} This proves that $f(x)$ is the only limit point of $\{f(x_n)\}$ which lies in the compact set $M$. Hence $f(x_n)\to f(x)$, this shows the continuity of $f$.

There're some points of the proof that I still yet to comprehend. First, I understand the existance of $f(x)$ due to the continuity of the function $y\longmapsto\|x-y\|$ on the compact $M$. However, I didn't see why $f(x)$ is unique.

Next, I don't understand why from (1) we got (2). In concrete, I don't get why we assumed that $a$ is the limit of $f(x_n)$. What if $f(x_n)$ isn't convergent ?

I really hope someone would help me to explain it more explicitly. Thanks.

Answer 1

That $f(x)$ is unique follows since $M$ is convex, a proof can be found here: closed convex set, unique point that minimizes distance.

There is a funky theorem in real analysis which states that if every subsequence of $(x_n)$ has a further subsequence that converges to $x$ (the same $x$ for all subsequences!), then $x_n \to x$ (Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.). If I'm not mistaken this result should give us what we want.

Take a subsequence $(f(x_{n_k}))$. To emphasise dependence on the subsequence, let's say $\{n_k : k \in \mathbb N\} = I$. Since $(f(x_{n_k}))$ is valued in $M$, it is a bounded sequence and so by Bolzano-Weierstrass it has a convergent subsequence $(f(x_{n_{k_t}}))$ with limit $a_I$. Since $M$ is closed, this limit will be in $M$. As the proof says, we will have $\lVert x - a_I\rVert \le \lVert x - y\rVert$ for all $y \in M$, and hence $a_I$ is the unique minimiser $a = f(x)$. Since every subsequence of $(f(x_n))$ has a further subsequence converging to $f(x)$, we have $f(x_n) \to f(x)$. Notice that the uniqueness of the minimiser is crucial - in general there is no reason to expect all the limits $a_I$ to coincide.

This might be what the author was shooting for, but I am not convinced by what they wrote. Maybe I am missing something simpler.

Answer 2

Once we prove that $f$ is well-defined (that is, that the $\arg\min$ exists and is unique), here is a different, geometric argument to show that it's not only continuous, but a weak contraction: it does not increase the distance between any two points. This is in some way a generalization of the property that projection onto a line is a weak contraction, which I'll use in the proof below.

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Let $x$ and $x'$ be any two points in $\mathbb R^n$; let $y = f(x)$ and $y' = f(x')$. I will assume $y \ne y'$, since otherwise, we're done.

Let $\ell$ be the line through $y$ and $y'$, and let $p$ be the projection of $x$ onto $\ell$. The points $p, y, y'$ must occur in that order on $\ell$, possibly with $p=y$. To see this, I'll rule out the other possibilities:

• $p$ cannot be strictly between $y$ and $y'$ because the entire segment $[y,y']$ is contained in $M$ (by convexity) and this would make $p$ a point in $M$ closer to $x$ than $y$ is.
• $p$ cannot be closer to $y'$ than to $y$, because then we'd have $$\|x-y'\|^2 = \|x-p\|^2 + \|p-y'\|^2 < \|x-p\|^2 + \|p-y\|^2 = \|x-y\|^2,$$ and $y'$ would be closer to $x$ than $y$ is.

Similarly, if $p'$ is the projection of $x'$ onto $\ell$, then the points $y, y', p'$ must occur in that order on $\ell$, possibly with $p' = y'$. That is, the line $\ell$ looks something like: $$\longleftarrow p - y - y'- p' \longrightarrow$$ In particular, $\|p-p'\| \ge \|y-y'\|$. But we also have $\|x-x'\| \ge \|p-p'\|$ because projection onto $\ell$ is a weak contraction. Therefore $\|y-y'\| \le \|x-x'\|$, as desired.