Is $$f(x,y,z)=\begin{cases}
x^2y^2z^2\sin\left(\frac{1}{xyz}\right), & xyz\neq 0\\
0, &\text{else}\end{cases}$$
differentiable everywhere?
Here is my proof.
If $x,y,z\neq0$ then $f$ is differentiable as a composition of differentiable functions.
if $x=0$ and $y=u,z=v$ then:
\begin{align*}
\lim_{(x,y,z)\to(0,u,v)} \left|\frac{x^2y^2z^2\sin(\frac{1}{xyz})}{\sqrt{x^2+(y-u)^2+(z-v)^2}}\right|&\leq \lim_{(x,y,z)\to(0,u,v)} |y^2z^2|\left|\frac{x^2}{\sqrt{x^2+(y-u)^2+(z-v)^2}}\right|\\[5pt]
&\leq u^2v^2 \lim_{(x,y,z)\to(0,u,v)} \left|\frac{x^2}{\sqrt{x^2}}\right|\\
&=0
\end{align*}
and so $f=0+o(|(x,y,z)-(0,u,v)|)$ so $Df=0$ and $f$ is differentiable.
In the same way we can show that if $x,y=0,z=u\neq0$ then $f$ is differentiable at $(0,0,u)$ with differential $0$.
Also at $(x,y,z)=(0,0,0)$ the function over $|(x,y,z)-(0,0,0)|$ is bounded by $C\cdot r^5$ and so its differentiable at $(0,0,0)$ ans all in all its differentiable everywhere.
Is my solution or am I missing something? I think I am correct, although my multivariable calculus is a bit rusty hence the post.
Best Answer
Your proof is right in principle (I didn’t read too carefully) but it can be much simpler. This is the composition $f\circ g$ of the differentiable function $$ f:\mathbb R\to \mathbb R, \quad f(r)=\begin{cases}r^2 \sin1/r&r\neq0\\0&r=0\end{cases}$$ With the differentiable function $$ g:\mathbb R^3 \to \mathbb R,\quad g(x,y,z)=xyz$$ By the multivariate chain rule, $f\circ g$ is differentiable.