Smoothness properties of partial antiderivatives I'm wondering what can be said about the well-beha

rs450nigglix2

rs450nigglix2

Answered question

2022-05-28

Smoothness properties of partial antiderivatives
I'm wondering what can be said about the well-behavior (or lack thereof) of antiderivatives of multivariable functions.
For example, consider the function f ( x , y ) = 2 x y. If we take the antiderivative of this function with respect to x, we obtain F ( x , y ) = x 2 y + h ( y ) where h(y) is an arbitrary function of y. The curious thing is there are no restrictions on what h could be. h could be a well-behaved differentiable function, or it could be a nasty function like the Weierstrass function, which is differentiable nowhere, or even the completely discontinuous rational number detector h ( x ) = { 1 if  x  is rational 0 if  x  is irrational
In either case, F would not be differentiable with respect to y, but even so, the partial derivative F / x would be well-defined and will equal our original function f.
So obviously not every possible partial antiderivative will be differentiable with respect to the other variable, but clearly some are. If we choose h ( y ) = y, for example, then F is certainly differentiable with respect to y. In fact, as long as h is a differentiable function itself, F will be differentiable with respect to y.
My question is whether this holds true in general. If partial antiderivatives F exist for a multivariable function f with respect to one variable, is it always true that at least one such F will be differentiable with respect to the other variables?
I'm guessing the answer to this is No, but that raises the question of what requirements must we impose on the original function f to guarantee that a "nice" partial antiderivative exists. Is it enough that f be continuous? That f be fully differentiable? That f be continuously differentiable?
And even more generally, if I want to guarantee the existence of a partial antiderivative F that has some nice smoothness property (e.g. it is C 2 , or C 3 , etc.), what requirements must I impose on the original f to ensure this?

Answer & Explanation

ryancameron52

ryancameron52

Beginner2022-05-29Added 8 answers

Step 1
The naive thing to do is to define F by the formula
F ( x , y ) = 0 x f ( t , y ) d t .
Assuming f(⋅,y) is continuous for each fixed y, the fundamental theorem of calculus will guarantee that F x exists and equal to f. If in addition both f and f y are continuous, the differentiation under the integral sign theorem will guarantee that F y exists and is equal to 0 x f y ( t , y ) d t ..
Step 2
Under all those assumptions, both F and its partial derivatives are continuous so F is continuously differentiable. If you want F to be C k it is enough to require that f is C k . Those are sufficient conditions to impose on f to guarantee such F but they are by no means necessary.

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