If $f$ is a function from $\mathbb{R}^2$ to $\mathbb{R}$, not necessarily well-behaved. Then does the equation $$\frac{\partial f(t,u)}{\partial t}\Big|_{u=0}=\frac{\partial f(t,0)}{\partial t}$$ hold? Which means that the l.h.s. is defined if and only if the r.h.s is defined, and their values are equal in that case.
Does partial derivative and partial evaluation commute?
2
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calculus
multivariable-calculus
1 Answers
1
$$ f\left(t,0\right)=f\left(t,u\right)|_{u=0} $$
$$ \frac{\partial f\left(t,0\right)}{\partial t}=\lim_{\epsilon\to0}\frac{f\left(t+\epsilon,0\right)-f\left(t,0\right)}{\epsilon}\quad\mbox{(eq1)} $$
$$ \frac{\partial f\left(t,u\right)}{\partial t}=\lim_{\epsilon\to0}\frac{f\left(t+\epsilon,u\right)-f\left(t,u\right)}{\epsilon} $$
$$ \frac{\partial f\left(t,u\right)}{\partial t}|_{u=0}=\lim_{\epsilon\to0}\frac{f\left(t+\epsilon,0\right)-f\left(t,0\right)}{\epsilon}\quad\mbox{(eq2)} $$
So we can see eq1 and eq2 are the same; the only thing which can go wrong is if one of our expressions here was not defined.
I'd say there is some subtelty here. The expression $\frac{\partial f\left(t,u\right)}{\partial t}$ may not be defined as a function because at some point $u$ the derivative may not exist, in this case you could either say, depending on your definition, that:
(1) $\frac{\partial f\left(t,u\right)}{\partial t}$ is simply not defined; or
(2) that it is defined but over a smaller domain. If you take it to be defined over a smaller domain, then as long as it is defined at $u=0$, then the equality will hold.
In other words:
(1) if the partial derivative exists at $u=0$, and you allow $\frac{\partial f\left(t,u\right)}{\partial t}$ to be undefined where the limit does not exist (rather than just saying that function is not defined in general if there is some point where it is not defined) then the equality will hold.
(2) if you take $\frac{\partial f\left(t,u\right)}{\partial t}$ to be undefined if there is even one point where the limit does not exist, then you could have a situation where $\frac{\partial f\left(t,u\right)}{\partial t}$ is not defined as a function, even though $\frac{\partial f\left(t,0\right)}{\partial t}$ is defined (namely, this will happen if $f$ is not differentiable at some point where $u\neq0$ but is differentable on $u=0$)