Is there a smooth function in the unit ball with compact support?

[docnet]

Homework Statement

Is there a smooth function in the unit ball with compact support?

Relevant Equations

Functions of the form:
$\begin{cases}f(x,y,z) \geq 0 & \text{for}\quad ||x^2+y^2+z^2||<1 \\ 0 & \text{elsewhere}\end{cases}$

Is there a function that takes positive values only in the unit ball not including the boundary points defined by the set $\{x^2+y^2+z^2<1\}$, and $0$ everywhere else?

 

[Office_Shredder]

Have you done any work on it? It might help to start by thinking about this in 1 dimension.

 

[docnet]

So sorry, I forgot that one needs to attempt a solution.

In 1 dimension, I think a piece-wise polynomial function like
$$f(x)=\begin{cases}
0 & x<-1 \\ (x+1)^4 & -1\leq x < -\frac{1}{2} \\ 2x^4-\frac{3}{2}x^2+\frac{5}{16} & -\frac{1}{2} \leq x<\frac{1}{2} \\ (x-1)^4 & \frac{1}{2}\leq x<1 \\
0 & 1\leq x
\end{cases}$$
is ok. To extend this to the 3D case, maybe something like
$$g(x)=\begin{cases}
0 & |(x,y,z)|<-1 \\ (x+y+z+1)^4 & -1\leq |(x,y,z)| < -\frac{1}{2} \\ 2(x^4+y^4+z^4)-\frac{3}{2}(x^2+y^2+z^2)+\frac{5}{16} & -\frac{1}{2} \leq |(x,y,z)|<\frac{1}{2} \\ (x-y-z-1)^4 & \frac{1}{2}\leq |(x,y,z)|<1 \\
0 & 1\leq |(x,y,z)|
\end{cases}$$
is ok?

 

[Office_Shredder]

For the multi dimensional case try just making your 1d function a function of distance from the origin.

 

[Mark44]

This was definitely not a precalculus sort of question, so I have moved it to the calculus section. "smooth" implies properties of both the function and its derivative.

 

[Office_Shredder]

Oh, your 1d function also isn't smooth. The fourth derivative of $(x+1)^4$ is 24.

 

[Keith_McClary]

There are degrees of Smoothness .

 

[Office_Shredder]

There are degrees of Smoothness .

The function f is said to be[...] smooth [...]f it has derivatives of all orders

This is the normal definition of a function being smooth.

 

[docnet]

@Office_Shredder, @Keith_McClary, @Mark44, Thank you so much for your advice about the smoothness and making $f$ a function of distance from the origin. Unfortunately I am busy with other classes for a few days. I will work on this problem in a few days and come back when I believe I have it right.

 

[Orodruin]

The piecewise polynomials will never work out as they will not be smooth due to having different derivatives at at least some order (if not they are the same polynomial). For example, if your polynomials are of order four you cannot have different coefficients in front of the $x^4$ term because that would mean that the fourth derivative with respect to $x$ would not be continuous.

Also, examples can be found on the Wikipedia page already posted ...
See also Bump function.

 

[docnet]

@Orodruin thank you for your thoughtful advice regarding piecewise polynomials. It makes intuitive sense that they are only smooth up to a degree.

Per the wikipedia page, the bump function is $$f(x,y,z)=e^{-\frac{1}{1-x^2-y^2-z^2}}$$ or $$f(r)= e^{-\frac{1}{1-r^2}}$$ in polar coordinates

over the domain defined by the unit disk and $0$ else.

Here is a graph of the case when $f(x,y)=e^{-\frac{1}{1-x^2-y^2}}$

A confusing aspect of this assignment is my professor asks us to plot the graph of the case $f(x,y,z):\mathbb{R}^3\rightarrow \mathbb{R}$.

I wonder if he wants me to imagine an inventive way to illustrate the graph, explain why it is not possible, or maybe discover the $4^{th}$ spatial dimension using my portable pocket particle accelerator?

 

[Office_Shredder]

One interesting thing to note is that since all the derivatives at the boundary need to be 0, whatever function you are using inside the ball, it cannot be analytic (i.e. cannot be equal to its Taylor series locally at the boundary, since that Taylor series is identically zero).I have no idea how you are supposed to graph a 3d function.

 

[docnet]

One interesting thing to note is that since all the derivatives at the boundary need to be 0, whatever function you are using inside the ball, it cannot be analytic (i.e. cannot be equal to its Taylor series locally at the boundary, since that Taylor series is identically zero).I have no idea how you are supposed to graph a 3d function.

That is interesting and mysterious!