From Rudin's book, theorem 7.17: Suppose $f_n$ is a sequence of differentiable functions on $[a,b]$, and which converges at some point $x_0\in[a,b]$. If $g_n:=f_n'$ converges uniformly on $[a,b]$ then $f_n\to f$ uniformly for some $f$ which is differentiable and for all $x\in [a,b]$, $f'(x)=\lim\limits_{n\to\infty}g_n(x)$.
This statement only considers the derivative with respect to one variable.
My question is that how can it be generalised to higher number of variables. In particular:
Suppose I have a sequence of function $f_n(x,y,z,t)$, $g_n(x,y,z,t)$, with the relation that
$$
\forall n, \hspace{5mm} \frac{d}{dx}f_n(x,y,z,t) + \frac{d}{dy}f_n(x,y,z,t) = g_n(x,y,z,t)
$$
Also
$$
\lim_{n \rightarrow \infty} f_n(x,y,z,t) = f(x,y,z,t)
$$
We know that $f_n$ and $f$ are continuous functions. How can we say that
$$
\lim_{n \rightarrow \infty} g_n(x,y,z,t) = \frac{d}{dx}f(x,y,z,t) + \frac{d}{dy}f(x,y,z,t)
$$
Is it correct to say that if $\frac{d}{dx}f_n(x,y,z,t)$ and $ \frac{d}{dy}f_n(x,y,z,t)$ are uniformly convergence then the expression holds?
Another question is that, the statement in Rudin's book is for the interval $[a,b]$, can it be extended to $\mathbb{R}$?
Best Answer
I don't think you want a sum of the derivatives here. The natural generalization to $\mathbb{R}^n$ would involve the gradient. Indeed, suppose $f, f_n:\mathbb{R}^4\rightarrow \mathbb{R}$. Suppose further that for each variable, the convergence is uniform holding the other variables fixed.
Define $g_n:\mathbb{R}^4\rightarrow \mathbb{R}^4$ as \begin{equation} g_n(x,y,z,w):=\nabla f_n(x,y,z,w). \end{equation} Hence, \begin{equation} \begin{split} \lim_{n\rightarrow \infty}g_n(x,y,z,w)&=\lim_{n\rightarrow \infty}\nabla f_n(x,y,z,w)\\ &=\lim_{n\rightarrow \infty}\langle\frac{\partial f_n}{\partial x}, \frac{\partial f_n}{\partial y}, \frac{\partial f_n}{\partial z}, \frac{\partial f_n}{\partial w}\rangle \end{split} \end{equation}
Since the convergence of the $f_n$'s was uniform for each variable holding the other variables fixed, the partials also converge. Hence $\lim_{n\rightarrow \infty}g_n=\nabla f$.