# Non linear map preserving Euclidean norm

Let $$V$$ be a real vector space endowed with an Euclidean norm $$\Vert \cdot \Vert$$.

A bijective map $$T : V \to V$$ that preserves inner product $$\langle \cdot, \cdot \rangle$$ is linear. Also, Mazur-Ulam theorem states that an onto map $$T : V \to V$$ which is an isometry ($$\Vert T(x)-T(y) \Vert = \Vert x-y \Vert$$ for all $$x,y \in V$$) and fixes the origin ($$T(0) = 0$$) is linear.

What about an application that preserves the norm ($$\Vert T(x) \Vert = \Vert x \Vert$$ for all $$x \in V$$)? $$T$$ might not be linear as we show with following example:$\begin{array}{l|rcll} T : & V & \longrightarrow & V \\ & x & \longmapsto & x & \text{if } \Vert x \Vert \neq 1\\ & x & \longmapsto & -x & \text{if } \Vert x \Vert = 1\end{array}$

It is clear that $$T$$ preserves the norm. However $$T$$ is not linear as soon as $$V$$ is not the zero vector space. In that case, consider $$x_0$$ such that $$\Vert x_0 \Vert = 1$$. We have:$\begin{cases} T(2 x_0) &= 2 x_0 \text{ as } \Vert 2 x_0 \Vert = 2\\ \text{while}\\ T(x_0) + T(x_0) = -x_0 + (-x_0) &= – 2 x_0 \end{cases}$

# Non linear map preserving orthogonality

Let $$V$$ be a real vector space endowed with an inner product $$\langle \cdot, \cdot \rangle$$.

It is known that a bijective map $$T : V \to V$$ that preserves the inner product $$\langle \cdot, \cdot \rangle$$ is linear.

That might not be the case if $$T$$ is supposed to only preserve orthogonality. Let’s consider for $$V$$ the real plane $$\mathbb R^2$$ and the map $\begin{array}{l|rcll} T : & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (x,y) & \longmapsto & (x,y) & \text{for } xy \neq 0\\ & (x,0) & \longmapsto & (0,x)\\ & (0,y) & \longmapsto & (y,0) \end{array}$

The restriction of $$T$$ to the plane less the x-axis and the y-axis is the identity and therefore is bijective on this set. Moreover $$T$$ is a bijection from the x-axis onto the y-axis, and a bijection from the y-axis onto the x-axis. This proves that $$T$$ is bijective on the real plane.

$$T$$ preserves the orthogonality on the plane less x-axis and y-axis as it is the identity there. As $$T$$ swaps the x-axis and the y-axis, it also preserves orthogonality of the coordinate axes. However, $$T$$ is not linear as for non zero $$x \neq y$$ we have: $\begin{cases} T[(x,0) + (0,y)] = T[(x,y)] &= (x,y)\\ \text{while}\\ T[(x,0)] + T[(0,y)] = (0,x) + (y,0) &= (y,x) \end{cases}$

# A proper subspace without an orthogonal complement

We consider an inner product space $$V$$ over the field of real numbers $$\mathbb R$$. The inner product is denoted by $$\langle \cdot , \cdot \rangle$$.

When $$V$$ is a finite dimensional space, every proper subspace $$F \subset V$$ has an orthogonal complement $$F^\perp$$ with $$V = F \oplus F^\perp$$. This is no more true for infinite dimensional spaces and we present here an example.

Consider the space $$V=\mathcal C([0,1],\mathbb R)$$ of the continuous real functions defined on the segment $$[0,1]$$. The bilinear map
$\begin{array}{l|rcl} \langle \cdot , \cdot \rangle : & V \times V & \longrightarrow & \mathbb R \\ & (f,g) & \longmapsto & \langle f , g \rangle = \displaystyle \int_0^1 f(t)g(t) \, dt \end{array}$ is an inner product on $$V$$.

Let’s consider the proper subspace $$H = \{f \in V \, ; \, f(0)=0\}$$. $$H$$ is an hyperplane of $$V$$ as $$H$$ is the kernel of the linear form $$\varphi : f \mapsto f(0)$$ defined on $$V$$. $$H$$ is a proper subspace as $$\varphi$$ is not always vanishing. Let’s prove that $$H^\perp = \{0\}$$.

Take $$g \in H^\perp$$. By definition of $$H^\perp$$ we have $$\int_0^1 f(t) g(t) \, dt = 0$$ for all $$f \in H$$. In particular the function $$h : t \mapsto t g(t)$$ belongs to $$H$$. Hence
$0 = \langle h , g \rangle = \displaystyle \int_0^1 t g(t)g(t) \, dt$ The map $$t \mapsto t g^2(t)$$ is continuous, non-negative on $$[0,1]$$ and its integral on this segment vanishes. Hence $$t g^2(t)$$ is always vanishing on $$[0,1]$$, and $$g$$ is always vanishing on $$(0,1]$$. As $$g$$ is continuous, we finally get that $$g = 0$$.

$$H$$ doesn’t have an orthogonal complement.

Moreover we have
$(H^\perp)^\perp = \{0\}^\perp = V \neq H$