Cubic addition and differentiablility

It came to my thought that if we define $a\oplus b = (a^3 + b^3)^{\frac13}$ then $\Bbb R$ endowed with $\oplus$ and $\cdot$ the latter being the usual multiplication is a field, with usual $0$ and $1$ being the zero and the unity of this field respectively. Not really knowing what to do next with this, I decided to check how will new derivatives be different from the old ones. If I am not mistaken, it holds that $$ \lim_{h\to 0}\frac{f(x\oplus h)\ominus f(x)}{h} = \left(f'(x)\frac{f^2(x)}{x^2}\right)^{\frac13} $$ if $x\neq 0$, while the direct evaluation of the limit when $x = 0$ gives that indeed it is infinite unless $f(0) = 0$. In the latter case we get the new derivative (namely, the limit) being exactly the old derivative $f'(0)$. This looks very odd to me: why $0$ should be any special point of this field?

So my questions are:

1. Did I make a mistake somewhere? If yes, please point me to it.

2. If not, why in this new field differentiability at $0$ is something special?

Solution 1:

$\def\D{\mathscr{D}}\def\H{\mathscr{H}}\def\R{\mathbb{R}}\DeclareMathOperator{\id}{id}\DeclareMathOperator{\sgn}{sgn}\def\abs#1{\left|#1\right|}\def\paren#1{\left(#1\right)}$All the terms in this answer are in the classic sense unless specified otherwise.

Define\begin{align*} \H &= \{φ: \R → \R \mid φ\ \text{is an increasing homeomorphism},\ φ(0) = 0\},\\ \D_0 &= \{φ \in \H \mid φ\ \text{is derivable on}\ \R^*,\ φ'(x) ≠ 0\ (\forall x ≠ 0)\},\\ \D &= \{φ \in \H \mid φ\ \text{is a diffeomorphism}\}, \end{align*} and for any $φ \in \H$, define\begin{gather*} x +_φ y = φ^{-1}(φ(x) + φ(y)),\\ D_φ f(x) = \lim_{h → 0} \frac{1}{h} (f(x +_φ h) -_φ f(x)). \end{gather*} Note that the metric induced by $+_φ$, i.e.$$ d_φ(x, y) = \left\{ \begin{array}{ll} x -_φ y; & x \geqslant y\\ y -_φ x; & x < y \end{array} \right\} = φ^{-1}(|φ(x) - φ(y)|), $$ is equivalent to the classic metric $d(x, y) = |x - y|$, so $D_φ$ is well-defined.

It is natural to assume that $φ$ and $f$ are derivable to analyze $D_φ f$, but the cases where $φ \in \D$ yields the most intuitions. For $φ \in \D$ and any derivable $f$,\begin{align*} D_φ f(x) &= \left. \frac{\partial}{\partial h}(f(x +_φ h) -_φ f(x)) \right|_{h = 0}\\ &=\bigl( (φ^{-1})'(φ(f(x +_φ h)) - φ(f(x))) · φ'(f(x +_φ h)) · f'(x +_φ h) ·\\ &\mathrel{\phantom=} (φ^{-1})'(φ(x) + φ(h)) · φ'(h) \bigr)\Bigr|_{h = 0}\\ &= \color{red}{ \frac{1}{φ'(0)} } · φ'(f(x)) · f'(x) · \color{blue}{ \frac{1}{φ'(x)} } · \color{red}{ φ'(0) } \tag{1}\\ &= \frac{φ'(f(x)) f'(x)}{φ'(x)} = \frac{(φ \circ f)'(x)}{φ'(x)}, \end{align*} so $D_φ f = \dfrac{D(φ \circ f)}{Dφ}$, where $D$ is the classic derivative operator. Note that this formula also implies that diffeomorphism in the classic sense is equivalent to that in the sense of $D_φ$.

Now, in order to show that $(\R, +)$ cannot be differentiated from $(\R, +_φ)$ by how their derivative operators interact with other non-classic plus operations, define\begin{gather*} x +_{φ_1, φ_2} y = φ_2^{-1}(φ_2(x) +_{φ_1} φ_2(y)),\\ D_{φ_1, φ_2}f(x) = \lim_{h → 0} \frac{1}{h} (f(x +_{φ_1, φ_2} h) -_{φ_1, φ_2} f(x)) \end{gather*} for any $φ_1, φ_2 \in \D$. Since$$ x +_{φ_1, φ_2} y = φ_2^{-1}(φ_1^{-1}(φ_1(φ_2(x)) + φ_1(φ_2(y)))) = x +_{φ_1 \circ φ_2} y. $$ then $+_{φ_1, φ_2} = +_{φ_1 \circ φ_2}$ and$$ D_{φ_1, φ_2}f = D_{φ_1 \circ φ_2}f = \frac{D(φ_1 \circ φ_2 \circ f)}{D(φ_1 \circ φ_2)} = \frac{D(φ_1 \circ φ_2 \circ f) / Dφ_1}{D(φ_1 \circ φ_2) / Dφ_1} = \frac{D_{φ_1}(φ_2 \circ f)}{D_{φ_1}φ_2}. $$ This formula on $(\R, +_{φ_1})$ is just the same as $D_{\id, φ_2} f = \dfrac{D_{\id}(φ_2 \circ f)}{D_{\id} φ_2}$, or $D_{φ_2} f = \dfrac{D(φ_2 \circ f)}{Dφ_2}$, on $(\R, +)$.

The situation gets much more complicated even if the condition $φ \in \D$ is simply relaxed as $φ \in \D_0$. Now the definition of $\D_0$ ensures that the blue term in (1) is always defined for $x ≠ 0$, but the red terms suggest that the simple chain rule fails if $φ'(0) = 0$ or $φ'(0)$ does not exist. This suggests that $0$ is a special point for $D_φ f$.

More specifically, for $x ≠ 0$, since\begin{align*} &\mathrel{\phantom=} \left. \frac{\partial}{\partial k}φ(f(φ^{-1}(φ(x) + k))) \right|_{k = 0}\\ &= \bigl( φ'(f(φ^{-1}(φ(x) + k))) · f'(φ^{-1}(φ(x) + k)) · (φ^{-1})'(φ(x) + k) \bigr)\Bigr|_{k = 0}\\ &= φ'(f(x)) · f'(x) · \frac{1}{φ'(x)} = \frac{φ'(f(x)) f'(x)}{φ'(x)}, \end{align*} then\begin{gather*} \frac{1}{h} (f(x +_φ h) -_φ f(x)) = \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) }\ (h → 0). \tag{2} \end{gather*} Therefore $D_φf(x)$ exists if the limit of the RHS of (2) exists. Even though (2) is not valid for $x = 0$, the factor $φ'(f(x))$ in (2) implies that whether $f(0) = 0$ or $f(0) ≠ 0$ makes a difference for $D_φ f(0)$ if $φ'(0) = 0$ since there is a factor $φ'(x)$ in the denominator and derivatives are Darboux functions. Below are two explicit examples.

Example 1: $φ(x) = \sgn(x) |x|^a$ ($a > 0$, $a ≠ 1$). Since $φ$ is multiplicative, then for $x ≠ 0$, (2) implies that\begin{align*} D_φf(x) &= \lim_{h → 0} \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) }\\ &= \lim_{h → 0} φ^{-1}\paren{ \frac{1}{φ(h)} \paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } }\\ &= \lim_{h → 0} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} + o(1) }\\ &= φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} } = \sgn(f'(x)) \abs{ \frac{f(x)}{x} }^{\frac{a - 1}{a}} |f'(x)|^{\frac{1}{a}}. \end{align*} Again $0$ can be seen to be a special point because of the factor $|f(x)|^{\frac{a - 1}{a}}$ in the above expression.

Example 2: $φ(x) = \begin{cases} x\exp\paren{-\dfrac{1}{x^2}}; & x ≠ 0 \\ 0; & x = 0 \end{cases}$. Note that for any $c \in (0, 1)$, $φ(ch) = o(φ(h))$ ($h → 0$). For $x ≠ 0$, if $f(x) ≠ 0$ and $f'(x) ≠ 0$, then for any $ε \in (0, 1)$,$$ φ((1 - ε)h) = o(φ(h)),\quad φ(h) = o(φ((1 + ε)h)),\quad (h → 0) $$ thus there exists $δ > 0$ such that for $h \in (-δ, δ) \setminus \{0\}$,$$ |φ((1 - ε)h)| < \abs{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } < |φ((1 + ε)h)|,\\ $$ so$$ 1 - ε < \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } < 1 + ε. $$ Therefore (2) implies that$$ D_φf(x) = \lim_{h → 0} \frac{1}{h} φ^{-1}\paren{ \frac{φ'(f(x)) f'(x)}{φ'(x)} · φ(h) + o(φ(h)) } = 1. $$ If $f(x) = 0$, then\begin{align*} D_φf(x) &= \lim_{h → 0} \frac{1}{h} φ^{-1}(φ(f(φ^{-1}(φ(x) + φ(h)))) - φ(0))\\ &= \lim_{h → 0} \frac{1}{h} f(φ^{-1}(φ(x) + φ(h)))\\ &= \left. \frac{\partial}{\partial h}(f(φ^{-1}(φ(x) + φ(h)))) \right|_{h = 0}\\ &= \bigl( f'(φ^{-1}(φ(x) + φ(h))) · (φ^{-1})'(φ(x) + φ(h)) · φ'(h) \bigr)\Bigr|_{h = 0}\\ &= f'(x) · \frac{1}{φ'(x)} · φ'(0) = 0. \end{align*} This example shows that if $φ$ varies too slowly near $0$, then $D_φ$ will almost wipe out any information of the functions it applies to.

Solution 2:

Analysis of the Limit

When $x\ne0$, $$ \begin{align} \lim_{h\to0}\frac{\color{#C00}{(x\oplus h)}-\color{#090}{x}}{h^3} &=\lim_{h\to0}\frac{x\left(\color{#C00}{\left(1+\frac{h^3}{x^3}\right)^{1/3}}-\color{#090}{1}\right)}{h^3}\tag{1a}\\ &=\frac1{3x^2}\tag{1b} \end{align} $$ Explanation:
$\text{(1a)}$: definition of $\oplus$
$\text{(1b)}$: Binomial Theorem

Therefore, $$ \begin{align} \lim_{h\to0}\frac{f(x\oplus h)-f(x)}{h^3} &=\color{#C00}{\lim_{h\to0}\frac{f(x\oplus h)-f(x)}{(x\oplus h)-x}}\color{#090}{\lim_{h\to0}\frac{(x\oplus h)-x}{h^3}}\tag{2a}\\ &=\frac{\color{#C00}{f'(x)}}{\color{#090}{3x^2}}\tag{2b} \end{align} $$ Explanation:
$\text{(2a)}$: limit of a product
$\text{(2b)}$: definition of derivative and $(1)$

Thus, $$ \begin{align} \lim_{h\to0}\frac{f(x\oplus h)\ominus f(x)}{h} &=\lim_{h\to0}\left(\frac{f(x\oplus h)^3-f(x)^3}{h^3}\right)^{1/3}\tag{3a}\\ &=\left(\color{#C00}{\lim_{h\to0}{\frac{f(x\oplus h)^3-f(x)^3}{f(x\oplus h)-f(x)}}}\color{#090}{\lim_{h\to0}\frac{f(x\oplus h)-f(x)}{h^3}}\right)^{1/3}\tag{3b}\\ &=\left(\color{#C00}{3f(x)^2}\color{#090}{\frac{f'(x)}{3x^2}}\right)^{1/3}\tag{3c}\\ &=\left(f'(x)\frac{f(x)^2}{x^2}\right)^{1/3}\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: definition of $\ominus$
$\text{(3b)}$: limit of a product
$\text{(3c)}$: derivative of $x^3$ and $(2)$
$\text{(3d)}$: rearrange

$(3)$ is as stated in the question.

When $x=0$, there are three cases:

$\boldsymbol{f(0)=0}$: $$ \begin{align} \lim_{h\to0}\frac{f(x\oplus h)\ominus f(x)}{h} &=\lim_{h\to0}\left(\frac{f(h)^3-f(0)^3}{h^3}\right)^{1/3}\tag{4a}\\ &=\lim_{h\to0}\frac{f(h)}{h}\tag{4b}\\ &=\lim_{h\to0}\frac{f(h)-f(0)}{h}\tag{4c}\\[6pt] &=f'(0)\tag{4d} \end{align} $$ Explanation:
$\text{(4a)}$: $0\oplus h=h$ and apply $\ominus$
$\text{(4b)}$: apply $f(0)=0$ and simplify
$\text{(4c)}$: apply $f(0)=0$
$\text{(4d)}$: definition of derivative

$(4)$ is as stated in the question, but there is another case where the limit exists.

$\boldsymbol{f(0)\ne0}$, but $\boldsymbol{f'(0)=f''(0)=0}$ $$ \begin{align} \lim_{h\to0}\frac{f(x\oplus h)\ominus f(x)}{h} &=\lim_{h\to0}\left(\frac{f(h)^3-f(0)^3}{h^3}\right)^{1/3}\tag{5a}\\ &=\left(\lim_{h\to0}\frac{f(h)^3-f(0)^3}{f(h)-f(0)}\lim_{h\to0}\frac{f(h)-f(0)}{h^3}\right)^{1/3}\tag{5b}\\ &=\left(3f(0)^2\frac{f'''(0)}6\right)^{1/3}\tag{5c}\\ &=\left(\frac{f(0)^2f'''(0)}2\right)^{1/3}\tag{5d} \end{align} $$ Explanation:
$\text{(5a)}$: $0\oplus h=h$ and apply $\ominus$
$\text{(5b)}$: limit of a product
$\text{(5c)}$: apply L'Hôpital thrice
$\text{(5d)}$: simplify

$\boldsymbol{f(0)\ne0}$ and $\boldsymbol{f'(0)\ne0}$ or $\boldsymbol{f''(0)\ne0}$

Same argument as $(5)$, but one of the applications of L'Hôpital gives an infinite limit.

Answers to the Questions

1. Did I make a mistake somewhere? If yes, please point me to it.

Not really a mistake, but an omission. There is a case where $f(0)\ne0$ and yet the limit exists. One example would be $e^{x^3}$.

2. If not, why in this new field differentiability at 0 is something special?

The difficulty arises because the functions of which we are taking derivatives with cubic addition and subtraction use the standard addition and subtraction in computing their values. When differentiating a function near $0$, $f$ can carry the large changes from cubic addition, $(x\oplus\Delta x)-x$, near $0$ to $f(0)$. If $f(0)\not\approx0$, cubic subtraction, $(f+\Delta f)\ominus f$, cannot undo these large changes, resulting in a large derivative.

Consider how the argument to the function is incremented by cubic addition $$ \begin{align} (x\oplus\Delta x)-x &=\left(x^3+\Delta x^3\right)^{1/3}-x\tag{6a}\\[6pt] &=x\left(\left(1+\frac{\Delta x^3}{x^3}\right)^{1/3}-1\right)\tag{6b}\\ &\approx\frac13\frac{\Delta x^3}{x^2}\tag{6c} \end{align} $$ Thus, $\Delta f\approx f'\frac{\Delta x^3}{3x^2}$. This is big if $x$ is close to $0$.

Next, consider how the cubic difference measures the change in $f$ $$ \begin{align} (f+\Delta f)\ominus f &=\left((f+\Delta f)^3-f^3\right)^{1/3}\tag{7a}\\ &=f\left(\left(1+\frac{\Delta f}{f}\right)^3-1\right)^{1/3}\tag{7b}\\ &\approx\left(3f^2\Delta f\right)^{1/3}\tag{7c} \end{align} $$ Plugging the $\Delta f$ from $(6)$ into $(7)$ gives $\left(f'\frac{f^2}{x^2}\right)^{1/3}\Delta x$, which matches the formula in $(3)$ (which is good).

If $f(0)\not\approx0$, the $f(x)^2$ in $\text{(7c)}$ cannot compensate for $\frac1{x^2}$ in $\text{(6c)}$.