# Exchanging limit and derivative for four variables

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 $$\mathbb{R}$$?

• Can you precise what you mean by $\frac{d}{dx}f_n(x,y,z,t)$ and $\frac{d}{dy}f_n(x,y,z,t)$ are uniformly convergence. Do you mean as maps of four variables on $\mathbb R^4$? May 5, 2021 at 12:01
• I'm not quite sure about what I'm saying. It can be in $\mathbb{R}^4$, or if we fix $z$ and $t$ it can be in $\mathbb{R}^2$. May 5, 2021 at 12:24

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 $$$$g_n(x,y,z,w):=\nabla f_n(x,y,z,w).$$$$ Hence, $$$$\begin{split} \lim_{n\rightarrow \infty}g_n(x,y,z,w)&=\lim_{n\rightarrow \infty}\nabla f_n(x,y,z,w)\\ &=\lim_{n\rightarrow \infty}\langle\frac{\partial f_n}{\partial x}, \frac{\partial f_n}{\partial y}, \frac{\partial f_n}{\partial z}, \frac{\partial f_n}{\partial w}\rangle \end{split}$$$$
Since the convergence of the $$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$$.
• Here, you only assume that the convergence of $f_n$ is uniform, if we fix all but one variable? you mean we don't need the uniform convergence of $g_n$'s. May 6, 2021 at 9:09
• Ok, thanks. Can you please provide some references for that? because anything I've seen requires that $g_n$ converges uniformly. May 6, 2021 at 17:54
• I'm sorry I misspoke. The $g_n$'s do need to uniformly converge. May 6, 2021 at 18:17