Existence of homotopic map

Question

Given two (smooth) maps $ L, R : [0,1] \to [0,1] $ is there a smooth homotopy $ H : [0,1] \times [0,1] \to [0,1] $ where the parameter $s$ gives you maps $ H_s : [0,1] \to [0,1], H_s(x) = H(x,s) $ such that $ H_0 = L, H_1 = R? $

Answer 1

It seems that your question is motivated by the Lorenz curve $L(p)$ as indicated by your question Constructing Lorenz-like curves . I do not know much about economic analysis, but let me try to assemble some assumptions on the function $L$ which seem to be implicit in your question. If some of these assumptions is improper, you can omit it.

a) $L : [0,1] \to [0,1]$ is a smooth function such that $L(x) = L(1-x)$ for all $x \in [0,1]$ (symmetry with respect to the diagonal $y = 1 -x$).

b) $L$ is strictly increasing and $L(0) = 0, L(1) = 1$. Topologically it is therefore an orientation preserving homeomorphism and has an inverse $L^{-1}$. If we want that $L^{-1}$ is smooth, we must strengthen "strictly increasing" to the requirement that $L'(x) > 0$ for all $x \in [0,1]$. In that case $(L^{-1})'(x) = 1/L'(L^{-1}(x)) > 0$.

c) $L(x) < x$ for all $x \in (0,1)$.

The function $R$ is the reflection of $L$ in the diagonal $y = x$. This means that each point $(x,R(x))$ is the reflection of some point $(y,L(y))$, that is, we have $(R(x),x) = (y,L(y))$. This implies $y = R(x)$ and $L(R(x)) = x$ for all $x$. Therefore $R = L^{-1}$ which is again symmetric with respect to $y = 1 - x$. Instead of c) it satisfies

c') $R(x) > x$ for all $x \in (0,1)$.

It seems you are looking for a homotopy $H$ such that "the function $L$ smoothly transitions from itself to its inverse $R$ as the parameter is varied and as it crosses the line $y = x$". My interpretation is that you want all intermediate functions $H_s$ to satisfy a), b), and c) for $s < \frac{1}{2}$ resp. c') for $s > \frac{1}{2}$, whereas for $s = \frac{1}{2}$ it should be the identity.

Now let us consider the homotopy $H$ defined in Berci's answer and check what can be said about the levels $H_s$.

(1) $H_s$ symmetric with respect to $y = 1 - x$:

$H_s(1- x) = sR(1-x) + (1-s)L(1-x) = sR(x) + (1-s)L(x) = H_s(x)$.

(2) $H_s$ is strictly increasing with a smooth inverse $H_s^{-1}$ (which means $H'_s(x) > 0$ for all $x$):

We have $H'_s(x) = sL'(x) + (1-s)R'(x) > 0$ because both $L'(x), R'(x) > 0$.

(3) In general $H_s$ does not satisfy c) or c'). As $s$ varies, $H_s(x)$ moves with constant speed from $L(x)$ to $R(x)$. For each $x \in (0,1)$ we have $H_{s(x)}(x) = x$ where $s(x) = \frac{x - L(x)}{R(x) - L(x)}$. The function $s(x)$ is in general not constant which means that the graph of $H_s$ intersects the diagonal for $s \in (0,1)$. I guess this happens for any $L$, but I do not have a proof. However, look at the example pictured in your question to verify the intersection phenomenon.

But I can offer something else. A rotation by $-\frac{\pi}{4}$ transforms $L, R$ into functions $L^\ast, R^\ast : [0, \sqrt{2}] \to [0, \sqrt{2}]$ such that $R^\ast(x) = -L^\ast(x)$. Now define $H^\ast : [0, \sqrt{2}] \times I \to [0, \sqrt{2}], H^\ast(x,s) = sR^\ast(x) + (1-s)L^\ast(x)$. This homotopy is "symmetric" with respect to the $x$-axis and all intermediate functions $H^\ast_s$ have the same characteristic features as $L^\ast$ for $s < \frac{1}{2}$ resp. as $R^\ast$ for $s > \frac{1}{2}$, whereas for $s = \frac{1}{2}$ it is constant with value $0$. A backward rotation by $\frac{\pi}{4}$ transforms the homotopy $H^\ast$ into the desired homotopy $H$.

Good explicit formulae for $L^\ast, R^\ast$ are not available (similarly as you do not have an explicit formula for $R = L^{-1}$ in terms of $L$). But here is how we can proceed. A rotation by $-\frac{\pi}{4}$ is described by the matrix

$\begin{bmatrix}\cos(-\frac{\pi}{4}) & -\sin(-\frac{\pi}{4})\\ \sin(-\frac{\pi}{4}) & \cos(-\frac{\pi}{4}) \end{bmatrix} = \begin{bmatrix}\cos(\frac{\pi}{4}) & \sin(\frac{\pi}{4})\\ -\sin(\frac{\pi}{4}) & \cos(\frac{\pi}{4}) \end{bmatrix} = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}$

Hence the point $(x,L(x)$ is rotated to $(\frac{1}{\sqrt{2}}(x + L(x)), \frac{1}{\sqrt{2}}(-x + L(x)) )$ which must have the form $(y,L^\ast(y))$. Define functions

$$L_\pm(x) = \frac{1}{\sqrt{2}}(\pm x + L(x)).$$

Then $L'_+(x) = \frac{1}{\sqrt{2}}(1 + L'(x)) > 0$, thus $L_+$ is a strictly increasing diffeomorphism $[0, \sqrt{2}] \to [0, \sqrt{2}]$. We see that $y = L_+(x)$ and $L^\ast(y) = L_-(x)$, therefore

$$L^\ast(y) = L_-(L_+^{-1}(y)) .$$

Similarly we construct $R^\ast$ and the backward rotation of $H^\ast$.

Answer 2

Yes: just connect up the corresponding points by line segments: $$H(x,s):=s\cdot R(x)+(1-s)\cdot L(x)$$ It is clearly smooth if $L,G$ are.

By the way, $[0,1]$ is contractible, meaning that any two continuous function to $[0,1]$ are homotopic to each other.