Let $U$ be an open set on ${\mathbb R}^{n}$ (but $U$ is not an empty set), $\textbf{p}\in{U}$, and $f:U\to \mathbb R$ is continuously-differentiable on $U$. Then, it is known that, "the function $f$ can be differentiable for all $\textbf{q}\in U$." (See Spivac)
And I know that, there is a function $f$ such that it is differentiable at $\textbf{p}$ but, for any $r> 0$, $f$ is not differentiable (and continuously differentiable) on
$U_{\textbf{p}} (r)$ .
Here, $U_{\textbf{p}} (r)$ is an open ball of radius $r$ centered on
$\textbf{p}$.
For example, if $f:{\mathbb R}^{2}\to \mathbb R$ is defined as follows, $f$ is differentiable at $\textbf{0}$,, but is not differentiable (and not continuous) at any other point. Here $\mathbb Q$ is the set of all rational numbers, and $U_{\textbf{p}} (r)$ is an open ball of radius $r$ centered on $\textbf{p}$.
$f(x,y):=\left\{
\begin{array}{rr}
0, & (x,y)\in \mathbb Q^{2} \\
x^2 + y^2, & (x,y)\notin \mathbb Q^{2} \\
\end{array}
\right.$
Therefore, there is at least one function that does not have a continuously-differentiable region, even if it can be differentiable at one point.
But I cannot imagine whether are there any functions that are differentiable on $U$ but not continuously-differentiable.
My question
Let $U$ be an open set of $\mathbb R^n$ (but is not an empty set), and $\ \textbf{p}\in U $.
Then, are there any functions $f:U\to \mathbb R$ such that,
$f$ is differentiable on $U$, but for any $r> 0$, $f$ is not continuously-differentiable on $U_{\textbf{p}} (r)$ ?
If so, give an example. If not, please explain why.
Here, $U_{\textbf{p}} (r)$ is an open ball of radius $r$ centered on $\textbf{p}$.
Here, the definitions of differentiable and continuously differentiable are as follows.
Def1 (Differentiable at $\textbf{p}$)
Let $U$ be an open set (but not empty set) of ${\mathbb R}^{n}$, $\textbf {p} \in \mathbb R^n$, and $f$ is a function whose domain is $U$.
At this time, $f$ is differentiable at $\textbf{p}$ iff the following is satisfied.
${\exists} A:{\mathbb R}^{n}\to \mathbb R$: a linear map such that
$${\lim}_{\textbf{x}\to\textbf{p}}\frac{|f(\textbf{x}) – A(\textbf{x}-\textbf{p}) – f(\textbf{p})|}{|\textbf{x}-\textbf{p}|} = 0$$
Def2 (Differentiable on $\textbf{U}$)
Let $U$ be an open set (but not empty sets) of ${\mathbb R}^{n}$, and $f$ is a function which domain is $U$.
At this time, $f$ is differentiable at $U$ iff "for all $\textbf{q}\in{\mathbb R}^{n}$, $f$ is differentiable at $\textbf{q}$".
Def3 (Continuously-differentiable on $U$)
Let $U$ be an open set (but is not empty sets) of ${\mathbb R}^{n}$, and $f$ is a function which domain is $U$.
At this time, $f:U\to \mathbb R$ is continuously-differentiable on $U$ iff
- $f$ is partially differentiable for all direction, ${x}_{1},
{x}_{2}, …, {x}_{n}$ (that mean, we can define $\frac{\partial f}{\partial{x}_{1}}, \cdots\frac{\partial f} {\partial{x}_{n}} $ on
$U$). and,- $\frac{\partial f} {\partial{x}_{1}}, \cdots\frac{\partial f}
{\partial{x}_{n}} $ are continuous on $U$.
P.S.
I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions.
Post-hoc Note:【Verification of the function taught by Thomas Shelby】
The following are the confirmation that the following function $f$ meets my requirement (Is it correct as proof?):
$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac1 {\sqrt{x^2+y^2}}\right),&(x,y)\neq 0\\0,&(x,y)=0\end{cases}.$
My proof:
${\lim}_{\|\textbf{x}\|\to 0} \frac{f(\textbf{x}) – f(\textbf{0})}{\|\textbf{x}\|}=
{\lim}_{\|\textbf{x}\|\to 0} \frac{{\|\textbf{x}\|}^{2}\sin(1/ \|\textbf{x}\| – 0)}{\|\textbf{x}\|}=
$
${\lim}_{\|\textbf{x}\|\to 0} \|\textbf{x}\|\sin(1/ \|\textbf{x}\|)
= 0$
Therefore, the $f$ is differntiable at $(0,0)$ and $Jf(0,0)=(0,0)$.On the other hand, for $\textbf{x}\neq\textbf{0}$,
Let $g$ and $h$ be $g(x,y):=\sqrt{{x}^2 + {y}^2}\ $ and $\ h(t):={t}^{2}\sin(1/t)$ (for $t\neq 0$) respectively, then$$\frac{d\sqrt{t}}{dt} = \frac{1}{2\sqrt{t}} $$ and,
$$(J\|\textbf{x}\|^2)(x,y) = (2x,2y) ,$$ Therefore,
$$(Jg)(x,y) = \left(\frac{x}{\|\textbf{x}\|} , \frac{y}{\|\textbf{x}\|}\right)\quad (\textrm{for all $\textbf{x}\neq\textbf{0}\ $}),$$ and
$$\ \frac{d\sin(1/t)}{dt} = -\frac{\cos(1/t)}{t^2}\ \ \ (\textrm{at $t\neq 0$}).$$ Therefore,
$$\frac{dh}{dt} ={t}^{2}\frac{d\sin(1/t)}{dt} + 2t\sin(1/t) = -\cos(1/t) + 2t\sin(1/t).$$Therefore, at $\textbf{x}\neq \textbf{0}$,
$$Jf(x,y) = \left(\left.\frac{dh}{dt}\right|_{t=||\textbf{x}||}\right)(Jg)(x,y) = (-\cos(1/||\textbf{x}||) + 2t\sin(1/||\textbf{x}||))\left(\frac{x}{||\textbf{x}||} , \frac{y}{||\textbf{x}||}\right).$$Therefore,
$$\frac{\partial f}{\partial x} = -\frac{x\cos(1/||\textbf{x}||)}{||\textbf{x}||} + 2x\sin(1/||\textbf{x}||),\,\textbf{x}\neq\textbf{0}$$and
$$\frac{\partial f}{\partial y} = -\frac{y\cos(1/||\textbf{x}||)}{||\textbf{x}||} + 2y\sin(1/||\textbf{x}||),\,\textbf{x}\neq\textbf{0}.$$However, both $\dfrac{x\cos(1/||\textbf{x}||)}{||\textbf{x}||} $ and
$\dfrac{y\cos(1/||\textbf{x}||)}{||\textbf{x}||} $ do not get converted at $(0,0)$.So, both $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are not continuous at $(0,0)$. ■
Best Answer
An example for $n = 1$ from the theory of random walks. Let $f$ be a (-n everywhere) discontinuous Lebesgue measurable function on $\mathbb{R}$. Here's an example with $f$ bounded by $1$, just showing the part $x \in [-3,3]$. (Note that I have only barely subsampled the graph in this interval. If I were to fully sample it, this finite resolution representation would almost surely appear to be a solid rectangle of points of the graph. Actually produced by generating $10^6$ uniformly distributed reals in $[-1,1]$ assigned to evenly spaced abscissae, then plotting a subsample of size $10^4$.)
This function is almost surely nowhere continuous (as any open interval almost surely contains points of heights arbitrarily close to $-1$ and $1$). The integral of this function, $$ \int_{0}^x \; f(t) \,\mathrm{d}t $$ is differentiable, but there's no hope of continuous differentiability. Graph of the integral (actually, Riemann sum approximations using $10^6$ intervals in $[-3,3]$):
Picking a different instance of a bounded by $1$ discontinuous Lebesgue measurable function on $\mathbb{R}$ and integrating it the same way, we can graph the integral.
These are almost everywhere differentiable by construction (by Lebesgue's differentiation theorem); we know the derivative is $f$. (The theorem generalizes to $n > 1$ and the integral to $\int_{[0,x_1]\times [0,x_2] \times \cdots \times [0,x_n]} \; f(t) \,\mathrm{d}t$ where we understand the intervals to be $[0,a]$ when $0 \leq a$ and $[a,0]$ when $a < 0$.) In some way, "most" functions are everywhere discontinuous messes, so "most" functions can be integrated to a differentiable, but not continuously differentiable, function.
(This construction can be iterated to get a function that is several times continuously differentiable, but whose "last" derivative is not continuous.)