# Distance between a point and a hyperplane not reached

Let’s investigate the following question: “Is the distance between a point and a hyperplane always reached?”

In order to provide answers to the question, we consider a normed vector space $$(E, \Vert \cdot \Vert)$$ and a hyperplane $$H$$ of $$E$$. $$H$$ is the kernel of a non-zero linear form. Namely, $$H=\{x \in E \text{ | } u(x)=0\}$$.

## The case of finite dimensional vector spaces

When $$E$$ is of finite dimension, the distance $$d(a,H)=\inf\{\Vert h-a \Vert \text{ | } h \in H\}$$ between any point $$a \in E$$ and a hyperplane $$H$$ is reached at a point $$b \in H$$. The proof is rather simple. Consider a point $$c \in H$$. The set $$S = \{h \in H \text{ | } \Vert a- h \Vert \le \Vert a-c \Vert \}$$ is bounded as for $$h \in S$$ we have $$\Vert h \Vert \le \Vert a-c \Vert + \Vert a \Vert$$. $$S$$ is equal to $$D \cap H$$ where $$D$$ is the inverse image of the closed real segment $$[0,\Vert a-c \Vert]$$ by the continuous map $$f: x \mapsto \Vert a- x \Vert$$. Therefore $$D$$ is closed. $$H$$ is also closed as any linear subspace of a finite dimensional vector space. $$S$$ being the intersection of two closed subsets of $$E$$ is also closed. Hence $$S$$ is compact and the restriction of $$f$$ to $$S$$ reaches its infimum at some point $$b \in S \subset H$$ where $$d(a,H)=\Vert a-b \Vert$$.

## An initial consideration regarding infinite dimensional vector spaces

We now suppose that $$E$$ is of infinite dimension. We first notice that $$H$$ is not necessarily closed. One can prove that $$H$$ is closed if and only if $$u$$ is continuous. When $$u$$ is discontinuous, $$H$$ is dense in $$E$$, hence for any $$a \in E$$ we have $$d(a,H)=0$$.

For example, consider the space $$X$$ of real-valued smooth functions on the interval $$[0, 1]$$ with the uniform norm, that is:
$\Vert f \Vert = \sup\limits_{x \in [0,1]} \vert f(x) \vert$ The derivative-at-a-point map, given by $$T(f)=f^\prime(0)$$ defined on $$X$$ and with real values, is linear but not continuous. Indeed, consider the sequence $$\displaystyle f_n(x)=\frac{\sin(n^2x)}{n}$$ for $$n \ge 1$$. This sequence converges uniformly to the constantly zero function, but $$T(f_n)=n \to + \infty$$ as $$n \to + \infty$$ instead of $$T(f_n) \to T(0) =0$$ which would hold for a continuous map. For any $$f \in X$$, consider the sequence $$\displaystyle g_n(x)=f(x)-f^\prime(0)\frac{\sin(nx)}{n}$$ for $$n \ge 1$$. We have $$g_n^\prime(x)=f^\prime(x)-f^\prime(0) \cos(nx)$$ and therefore $$g_n^\prime(0)=0$$. This means that $$T(g_n)=0$$ or in other words that $$g_n$$ belongs to the hyperplane $$H \equiv T(f)=0$$. As the sequence $$f^\prime(0)\frac{\sin(nx)}{n}$$ converges uniformly to the constantly zero function we conclude that $$d(f,H)=0$$.

## The case of Hilbert spaces

We now suppose that $$E$$ is a Hilbert space and $$H$$ a closed hyperplane defined by the equation $$u(x)=0$$. $$H$$ is a non-empty convex subset of $$E$$. According to the Hilbert projection theorem for every $$a \in E$$ there exists a unique $$b \in H$$ for which $$\Vert a-b \Vert$$ is minimized over $$H$$.
The proof of the Hilbert projection theorem is based on the parallelogram identity.

## A counterexample in the Banach space $$c_0$$

Finally we look at a counterexample of a Banach space where the distance of a point to a closed hyperplane is not reached.
We take for $$E$$ the sequence space $$c_0$$ of real sequences converging to zero equipped with the supremum norm $$\Vert x \Vert = \sup\limits_n \vert x_n \vert$$. $$c_0$$ is a Banach space. The linear form:
$\begin{array}{l|rcl} u : & c_0 & \longrightarrow & \mathbb{R} \\ & x & \longmapsto & \displaystyle \sum_{n=0}^\infty 2^{-n}x_n \end{array}$ is continuous as for $$x \in c_0$$ the inequality $$\displaystyle \vert u(x) \vert \le (\sum_{n=0}^\infty 2^{-n}) \Vert x \Vert = 2 \Vert x \Vert$$ holds. Consequently the hyperplane $$H=u^{-1}(\{0\})$$ is closed. If we denote by $$\Vert l \Vert$$ the norm $$\Vert l \Vert=\sup\limits_{\Vert x \Vert =1} \vert l(x) \vert$$ of any continuous linear form $$l$$, we have $$\Vert u \Vert=2$$. The above inequality proves that $$\Vert u \Vert \le 2$$. For $$p \in \mathbb{N}$$ we also have $$u(e(p)) = 2(1-2^{-(p+1)})$$, where $$e(p)$$ is the real sequence (belonging to $$c_0$$) with elements $$e(p)_n$$ equal to $$1$$ for $$n \le p$$ and to $$0$$ for $$n > p$$. Hence $$\Vert u \Vert \ge 2(1-2^{-(p+1)})$$ for all $$p \in \mathbb{N}$$ and therefore:
$\Vert u \Vert = 2$ as desired. We now use the following general statement (G): for any continuous linear form $$u$$ and any point $$a$$ in a vector space $$E$$, we have $$\displaystyle d(a,H)=\frac{\vert u(a) \vert}{\Vert u \Vert}$$.

Let’s suppose that for $$a \in c_0 \setminus H$$ the distance $$d(a,H)$$ was reached at a point $$b \in H$$. Using the statement (G), we would have:
$\vert u(a) \vert = 2 \Vert a- b \Vert$ according to $$\Vert u \Vert = 2$$. As $$\displaystyle u(b) = \sum_{n=0}^\infty 2^{-n}b_n=0$$, we can reformulate above equality as:
$\left \vert \displaystyle \sum_{n=0}^\infty 2^{-n}(a_n-b_n) \right \vert = 2 \Vert a – b \Vert = 2 \sup\limits_n \vert a_n – b_n \vert$ which is possible only if $$\vert a_p – b_p \vert = \Vert a – b \Vert$$ for all $$p \in \mathbb{N}$$. Indeed, given any $$p \in \mathbb{N}$$ we have:
\begin{aligned} \left \vert \displaystyle \sum_{n=0}^\infty 2^{-n}(a_n-b_n) \right \vert &= \left \vert 2^{-p}(a_p-b_p) + \displaystyle \sum_{\substack{n=0 \\ n \neq p}}^\infty 2^{-n}(a_n-b_n) \right \vert\\ &\le 2^{-p}\vert a_p-b_p \vert + \left \vert \displaystyle \sum_{\substack{n=0 \\ n \neq p}}^\infty 2^{-n}(a_n-b_n) \right \vert\\ &\le 2^{-p} \vert a_p-b_p \vert + 2 \Vert a- b \Vert – 2^{-p} \Vert a- b \Vert\ \end{aligned} and the last right hand term can be equal to $$2 \Vert a – b \Vert$$ only if $$\vert a_p – b_p \vert = \Vert a – b \Vert$$. This leads to a contradiction: as $$a_p \to 0$$ and $$b_p \to 0$$ for $$p \to +\infty$$, we get $$\Vert a – b \Vert=0$$ which is not possible as we supposed $$a \notin H$$.

## Proof of the general statement (G)

Take any continuous linear form $$u$$ and $$a \in E$$. We can suppose that $$a \notin H$$ (where $$H$$ is the kernel of $$u$$) as the statement is trivial when $$a \in H$$. For any $$h \in H$$ the norm of the vector $$(a-h)/\Vert a – h \Vert$$ is equal to $$1$$. Hence by definition of $$\Vert u \Vert$$ we have:
$\left \vert u \left (\frac{a-h}{\Vert a-h\Vert} \right ) \right \vert = \frac{\vert u(a-h) \vert}{\Vert a-h\Vert}=\frac{\vert u(a) \vert}{\Vert a-h\Vert} \le \Vert u \Vert$ Therefore $$\Vert a-h \Vert \ge \vert u(a) \vert / \Vert u \Vert$$ for all $$h \in H$$ and $$d(a,H) \ge \vert u(a) \vert /\Vert u \Vert$$.
On the other hand for all $$\epsilon > 0$$ one can find a vector $$x$$ whose norm is equal to one with $$\vert u(x) \vert \ge \Vert u \Vert/(1+\epsilon)$$. For $$\lambda = u(x)/u(a)$$, the vector $$h= x- \lambda a$$ belongs to $$H$$ as $$u(h)=0$$ and the equality $$\displaystyle x=\lambda(\frac{1}{\lambda}h+a)$$ holds. As $$\Vert x \Vert = 1$$ we get:
$1=\frac{\vert u(x) \vert}{\vert u(a) \vert} \left \Vert \frac{1}{\lambda}h+a \right \Vert \ge \frac{\Vert u \Vert}{\vert u(a) \vert(1+\epsilon)} \left \Vert \frac{1}{\lambda}h+a \right \Vert$ which implies that $$\displaystyle d(a,H) \le \frac{\vert u(a) \vert}{\Vert u \Vert} (1 + \epsilon)$$ as $$\displaystyle \frac{1}{\lambda}h \in H$$.

As the inequality holds for $$\epsilon > 0$$ as small as we want, we get the conclusion $$\displaystyle d(a,H) \le \frac{\vert u(a) \vert}{\Vert u \Vert}$$ and the statement (G) is proven.