X- normed space, Y – proper subspace. There exists such y, that ||x-y|| = dist(x,Y).

Question

Let $Y$ be a finite-dimensional subspace of a normed space $X$. Show that for every $x \in X$ there exists $y \in Y$ such that $\left\Vert x-y \right\Vert = \operatorname{dist}(x,Y)$.

Answer 1

Using some "well known" facts, this is not so difficult to prove.

Also, we assume the coefficient field is $\mathbb R$ or $\mathbb C$. Let's denote this coefficient field by $\mathbb K$.

Our first "well known" fact is, that on a finite dimensional vector space over $\mathbb K$, all norms are equivalent (see, for example, here: http://mathonline.wikidot.com/equivalence-of-norms-in-a-finite-dimensional-linear-space).

One useful consequence of this fact is that in the normed space $Y$ with the norm $||\cdot||$ inherited from $X$, the Heine-Borel theorem (see, for example, here: https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem) is true, i. e. a subset $S \subseteq Y$ is compact iff it is $||\cdot||$-closed and $||\cdot||$-bounded.

Now let's fix $x \in X$. Then the function $$ \varphi: Y \rightarrow \mathbb R_+, \quad y \mapsto ||y - x|| $$ is continuous on $Y$ (see, for example, here: Is the distance function in a metric space (uniformly) continuous?).

Now we will try to find a minimum of $\varphi$. Observe that $\varphi(0) = ||x||$, and if $||y|| > 3 ||x||$ then we have, using the triangle inequality, $$ \varphi(y) = ||y - x|| \geq ||y|| - ||x|| > 2 ||x||. $$ If we denote by $$ \overline{B_Y(0, 3||x||)} = \{ y \in Y | ||y|| \leq 3||x|| \} $$ the closed $||\cdot||$-ball in $Y$ centered at $0$ with radius $3||x||$, then we can rephrase the preceding two observations as follows. First, if $y \not\in \overline{B_Y(0, 3||x||)}$, then $\varphi(y) > 2 ||x||$. And second, for $0 \in \overline{B_Y(0, 3||x||)}$, we have $\varphi(0) = ||x|| \leq 2 ||x||$ (note that we could have $x = 0$). This means, that if $\varphi$ attains a minimum on $Y$, this minimum must be attained at a point in $\overline{B_Y(0, 3||x||)}$.

Now, $\overline{B_Y(0, 3||x||)}$ is closed and bounded, hence compact (see above), and $\varphi$ is continuous (see above), and suddenly we recall yet another "well known" fact, namely that a continuous real-valued function on a compact set has a finite infimum and attains this infimum (see, for example, here: https://en.wikipedia.org/wiki/Extreme_value_theorem#Generalization_to_metric_and_topological_spaces).

So we have shown that there is a $y_0 \in \overline{B_Y(0, 3||x||)} \subseteq Y$ such that $$ ||y_0 - x|| = \varphi(y_0) = \inf_{y \in \overline{B_Y(0, 3||x||)}}\varphi(y) = \inf_{y \in Y}\varphi(y) = \inf_{y \in Y}||y - x|| = dist(x, Y), $$ which is exactly what we wanted to prove.