# 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 $$d(a,H)=\Vert a-b \Vert$$. Continue reading Distance between a point and a hyperplane not reached

# A function whose derivative at 0 is one but which is not increasing near 0

From the mean value theorem, a real function whose derivative is strictly positive at every point of an interval is strictly increasing. In particular, a continuously differentiable function $$f$$ defined in a non-degenerate interval $$I$$ with a strictly positive derivative at a point $$a$$ of the interval is strictly increasing near that point. For the proof, we just have to notice that as $$f^\prime$$ is continuous and $$f^\prime(a) > 0$$, $$f^\prime$$ is strictly positive within an interval $$J \subset I$$ containing $$a$$. By the mean value theorem, $$f$$ is strictly increasing on $$J$$.

We now suppose that $$f$$ is differentiable on an interval $$I$$ containing $$0$$ with $$f^\prime(0)>0$$. For $$x>0$$ sufficiently close to zero we have $$\displaystyle \frac{f(x)-f(0)}{x-0} > \frac{f^\prime(0)}{2}>0$$, hence $$f(x)>f(0)$$. But that doesn’t imply that $$f$$ is strictly increasing in a neighborhood of zero. Let’s prove it with a counterexample. Continue reading A function whose derivative at 0 is one but which is not increasing near 0

# Two subgroups whose product is not a subgroup

In this article, we consider a group $$G$$ and two subgroups $$H$$ and $$K$$. Let $$HK=\{hk \text{ | } h \in H, k \in K\}$$.

$$HK$$ is a subgroup of $$G$$ if and only if $$HK=KH$$. For the proof we first notice that if $$HK$$ is a subgroup of $$G$$ then it’s closed under inverses so $$HK = (HK)^{-1} = K^{-1}H^{-1} = KH$$. Conversely if $$HK = KH$$ then take $$hk$$, $$h^\prime k^\prime \in HK$$. Then $$(hk)(h^\prime k^\prime)^{-1} = hk(k^\prime)^{-1}(h^\prime)^{-1}$$. Since $$HK = KH$$ we can rewrite $$k(k^\prime)^{-1}(h^\prime)^{-1}$$ as $$h^{\prime \prime}k^{\prime \prime}$$ for some new $$h^{\prime \prime} \in H$$, $$k^{\prime \prime} \in K$$. So $$(hk)(h^\prime k^\prime)^{-1}=hh^{\prime \prime}k^{\prime \prime}$$ which is in $$HK$$. This verifies that $$HK$$ is a subgroup. Continue reading Two subgroups whose product is not a subgroup