Let ##f(x,y)## be a scalar function. Then $$\frac{∂f}{∂x} = \lim_{h \rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h} = f_x (x,y)$$ and $$\frac{∂}{∂y} \left (\frac{∂f}{∂x} \right ) = \lim_{k \rightarrow 0} \frac{f_x(x,y+k)-f_x(x,y)}{k} = \lim_{k \rightarrow 0} \left ( \frac{ \displaystyle \lim_{h \rightarrow 0} \frac{f(x+h,y+k)-f(x,y+k)}{h} - \displaystyle \lim_{h \rightarrow 0} \frac{f(x+h,y)-f(x,y)}{h}} {k} \right ) = f_{yx} (x,y)$$(adsbygoogle = window.adsbygoogle || []).push({});

Similarly, $$\frac{∂f}{∂y} = \lim_{k \rightarrow 0} \frac{f(x,y+k)-f(x,y)}{k} = f_y (x,y)$$

and $$\frac{∂}{∂x} \left (\frac{∂f}{∂y} \right ) = \lim_{h \rightarrow 0} \frac{f_y(x+h,y)-f_y(x,y)}{h} = \lim_{h \rightarrow 0} \left ( \frac{ \displaystyle \lim_{k \rightarrow 0} \frac{f(x+h,y+k)-f(x+h,y)}{k} - \displaystyle \lim_{k \rightarrow 0} \frac{f(x,y+k)-f(x,y)}{k}} {h} \right ) = f_{xy} (x,y)$$

Now if we use the fact that $$\lim_{\mu \rightarrow \sigma} (g(\mu)) ± \lim_{\mu \rightarrow \sigma} ((h(\mu)) = \lim_{\mu \rightarrow \sigma}

(g(\mu) ± h(\mu))$$

then $$f_{yx} (x,y)= \lim_{k \rightarrow 0} \left ( \lim_{h \rightarrow 0} \left ( \frac{f(x+h,y+k)-f(x,y+k) - f(x+h,y) + f(x,y)}{hk} \right ) \right )$$ and $$f_{xy} (x,y) = \lim_{h \rightarrow 0} \left ( \lim_{k \rightarrow 0} \left ( \frac{f(x+h,y+k)-f(x+h,y) - f(x,y+k) + f(x,y)} {hk} \right ) \right ) $$

I'm just one short step from establishing the symmetry here (since the expression in both cases is the same), but I'm stuck. How do I justify interchanging the order of the limits? I don't think the answers obtained by applying the limits in a different order should in general commute.

Additionally, it is quite easy to see from above that this proof will be valid only if all the mentioned partial derivatives are defined (namely ##f_x(x,y)##,## f_y(x,y)##, ## f_{xy}(x,y)## and ##f_{yx} (x,y)## ), which is one of the required conditions for ##f_{xy} (x,y)## to be equal to ## f_{yx} (x,y)##. However, I can't find a justification for the other necessary condition - that the functions must be continuous.

Finally, I want to make sure this proof is formal (although it is admittedly very trivial in nature), so let me know if there's any step which seems a little hard to digest mathematically.

Any help is appreciated.

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# Proving equality of mixed second order partial derivatives

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