Exercise 15.2.5

Proof. (a) To apply the Chain Rule, we suppose that $g$ is continuously differentiable ( $g\in C_1({ℝ}^2)$). Write $x,y:{ℝ}^2\to {ℝ}^2$ the two maps defined by

Then

and

The Chain Rule gives

Suppose that $g(x+y,0)=g(x,y)$ for all $x,y\in ℝ$. Write $f(x)=g(x,0)$. Then $f$ is continuously differentiable, and $g(x,y)=f(x+y)$. By the Chain Rule, for all $(x,y)\in {ℝ}^2$,

therefore $\frac{\mathit{\partial g}}{\mathit{\partial x}}=\frac{\mathit{\partial g}}{\mathit{\partial y}}$ on ${ℝ}^2$.

Conversely, suppose that $\frac{\mathit{\partial g}}{\mathit{\partial x}}=\frac{\mathit{\partial g}}{\mathit{\partial y}}$ on ${ℝ}^2$. Then, for all $(u,v)\in {ℝ}^2$,

This means that for every fixed $u_0\in ℝ$, the map $v\mapsto h(u_0,v)$ has a null derivative, thus is constant: $h(u_0,v)=h(u_0,0)$ for all $v\in ℝ$. Since this is true for every $u_0$, we obtain

Write $f(u)=h(u,0)$ for all $u\in ℝ$. Then $f$ is continuously differentiable, and for all $u,v\in ℝ$, $h(u,v)=f(u)$ depends only of $u$.

By definition of $h$, this means that, for all $u,v\in ℝ$,

Taking $v=u$ in $g\left(\frac{1}{2}(u+v),\frac{1}{2}(u-v)\right)=h(u,v)=h(u,0)$, we obtain $g(u,0)=h(u,u)=h(u,0)$, therefore

thus

If $(x,y)$ is any pair in ${ℝ}^2$, there exists a unique pair $(u,v)\in {ℝ}^2$ such that $x=\frac{1}{2}(u+v),y=\frac{1}{2}(u-v)$, given by $u=x+y,v=x-y$. Therefore, the preceding equality implies that

(c) Define $g:{ℝ}^2\to ℝ$ by

The partial derivative of this quotient relative to the variable $x$ gives, using ${\phi }^{″}(x)=-2{\phi }^3(x)$ (see Exercise 3, part (d)), and ${\phi }^{\prime }{(x)}^2=1-{\phi }^4(x)$

This last expression is symmetric relatively to $x,y$, and also the denominator ${(1+{\phi }^2(x){\phi }^2(y))}^2$. Since $g(x,y)=g(y,x)=\frac{\phi (y){\phi }^{\prime }(x)+\phi (x){\phi }^{\prime }(y)}{1+{\phi }^2(y){\phi }^2(x)}$, this proves that

where $1+{\phi }^2(y){\phi }^2(x)>0$. Therefore $\frac{\mathit{\partial g}}{\mathit{\partial x}}=\frac{\mathit{\partial g}}{\mathit{\partial y}}$ on ${ℝ}^2$.

By part (b), $g(x,y)=g(x+y,0)$. Using $\phi (0)=0$, and ${\phi }^{\prime }(0)=\sqrt{1-{\phi }^4(0)}=1$,

We have proved the addition law for $\phi$:

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