# Differentiability of a function, of its sections and of its components

Let's consider two cases:

1. For $f:\mathbb{R}^2 \to \mathbb{R}$, where the domain may be some open subset in $\mathbb{R}^2$, define its sections to be functions in $\{ f(,x_2), f(x_1,), \forall x_1,x_2 \in \mathbb{R}\}$.

Do differentiablility of $f$ and differentiability of each of its sections in imply each other, or just one-directional implication, or neither?

2. For $f:\mathbb{R} \to \mathbb{R}^2$, where the domain may be some open subset in $\mathbb{R}$, define its components to be functions $\{f_i:\mathbb{R} \to \mathbb{R}, i=1,2, \text{ s.t. }f(x)=[f_1(x), f_2(x)], \forall x \in \mathbb{R} \}$.

Do differentiablility of $f$ and differentiability of each of its components imply each other, or just one-directional implication, or neither?

3. Are answers to part 1 and part 2 the same when $f:\mathbb{R}^n \to \mathbb{R}$ and $f:\mathbb{R} \to \mathbb{R}^n$?

Thanks and regards!

PS: In part 1, I was hoping to ask for a more general case $f:\mathbb{R}^n \to \mathbb{R}$, but didn't know how to write down a single formula for defining all its sections. Its sections are defined by fixing all except one of the n components of the domain variable and varying the exceptional one. Do you have some idea?

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What do you mean by "a single formula for defining all its sections"? – Jonas Meyer Jan 14 '12 at 18:44
I was sloppy. I meant, for example, a formula with a variable $i$, and then let $i$ vary from $1$ to $n$. That may be counted as more than one formulas. Also, if there are some notations for simplifying the definition? – Tim Jan 14 '12 at 18:49
Related page, where part 1. is answered. – Did Jan 14 '12 at 19:00

$\mathbb R^n\to\mathbb R$: Differentiability implies that the sections are differentiable.
A composition of differentiable functions is differentiable, and each of the functions $x\mapsto (x,x_2,x_3,\ldots)$, $x\mapsto (x_1,x,x_3,\ldots)$, etc., are differentiable. The sections of $f$ are compositions of $f$ with these functions.
The converse is false. A counterexample is the function $f:\mathbb R^2\to\mathbb R$ defined by $f(x,y)=\frac{xy}{x^2+y^2}$ if $x^2+y^2\neq 0$, $f(0,0)=0$. Each section of $f$ is differentiable everywhere, but $f$ is not even continuous at $(0,0)$, let alone differentiable there.
$\mathbb R\to\mathbb R^n$: Differentiability is equivalent to each component being differentiable. This is because convergence in $\mathbb R^n$ is equivalent to convergence in each component, and this can be applied to limits of difference quotients.
I don't know the best notation, but a section can be described as follows. Let $k\in\{1,2,\ldots,n\}$ and $x_1,x_2,\ldots,x_{k-1},x_{k+1},\ldots,x_n\in\mathbb R$ be fixed, and let $g:\mathbb R\to\mathbb R^n$ be defined by $g(x)=(x_1,x_2,\ldots,x_{k-1},x,x_{k+1},\ldots,x_n)$. Then $f\circ g$ is a section. – Jonas Meyer Jan 14 '12 at 18:36