# Proof for the "Fundamental Calculus Theorem" for two variables.

I`ve been stuck with this proof and mostly because I'm not sure if I can do a certain step, but at least without formality, it sounds good.

The problem goes like this: Be f $\in$C(R) such that $R=[a,b]\times[c,d]$, and we define $F(x,y)=\int^{x}_{a}\int^{y}_{c}f(u,v)du\cdot dv$. Prove that $\dfrac{\partial^{2}F}{\partial x \partial y}=\dfrac{\partial^{2}F}{\partial y \partial x}=f(x,y)$

Now the far that I got, formaly speaking, is to show that by taking the function f with an $x_o \in[a,b]$ fixed and then seeing that function like a composition of fuctions such that $f(h(y))=g(y)$ where $h(y)=x_o\widehat e_{x}+y\widehat e_{y}$, then I could apply the first part of the fundamental calculus theorem, and show that exist $G(y)=\int^y_cg(v)dv$ with the propety that $G'(y)=g(y)=f(x_o,y)$. Here is were my intuition takes place but I'm stuck trying to write it. Because this contruction leds to verify the existance of $G(y)$ for every $x\in[a,b]$, then I want to define $M(x)=\int^d_cH(y)du=\int^d_c[\int^b_af(u,v)dv]du$ and show that its derivative is $H(y)$ (I suppose that I could use the Fubini's Theorem to show the existance of $M(x)$ with the benefit that would ensure $M's$ integrability in $[a,b]$. But because Fubini's theorem shows the existance of that function with integral intervals from a to b, Im not sure if that could apply to my definition of $M$). And finally with all of this, having the existance of $\dfrac{\partial F}{\partial x},\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial^2F}{\partial x \partial y}$ being continuous by hypothesis, then by another theorem this implies that $\dfrac{\partial^{2}F}{\partial x \partial y}=\dfrac{\partial^{2}F}{\partial y \partial x}$.

Well this is more intuitive than formal, I would aprecciate some comments from my idea, NOT a solution, or something that could let me advance with the proof.

UPDATE: Justification for the iteration integral. Here is because it describes the function $$\frac{\partial}{\partial y}F(x,y) = \frac{\partial}{\partial y}\int_a^xG(u,y) \, du = \lim_{h\rightarrow 0}\dfrac{\int_a^xG(u,y+h)du-\int_a^xG(u,y)du}{h}$$ $$=\lim_{h\rightarrow 0}\int_a^x\dfrac{(G(u,y+h)-G(u,y))}{h}du$$ and because of the continuity of $G(u,y)$ in $[c,d]$ and its derivate on $y$ exists (for TFC on f) such that $$\dfrac{\partial}{\partial y}G(u,y)=f(u,y)$$ and then exist ($\int_a^b \dfrac{\partial}{\partial y}G(x,y) dx$) for the integrability on $f$, then applying the mean value theorem exists $\xi\in[y,y+h]$ such that

$$\left|\int_a^b \frac{G(x,y+h) - G(x,y)}{h} \, dx - \int_a^b \dfrac{\partial}{\partial y}G(x,y) \, dx\right|\\ = \left|\int_a^b \dfrac{\partial}{\partial y} G(x, \xi) \, dx - \int_a^b \dfrac{\partial}{\partial y}G(x,y) \, dx\right|\\ \leqslant \int_a^b |\dfrac{\partial}{\partial y}G(x,\xi) - \dfrac{\partial}{\partial y}G(x,y)| \, dx.$$

Since $f$ is continuous on $[a,b] \times [c,d]$, for its uniformly continuity, for any $\epsilon > 0$ there exists $\delta >0$ such that if $|h| < \delta$ then $|\dfrac{\partial}{\partial y}G(x,\xi) - \dfrac{\partial}{\partial y}G(x,y)|= |f(x,\xi) - f(x,y)|< \epsilon/(b-a)$ and

$$\int_a^b |\dfrac{\partial}{\partial y}G(x,\xi) - \dfrac{\partial}{\partial y}G(x,y)| \, dx < \epsilon.$$

Thus,

$$\lim_{h \rightarrow 0} \int_a^b \frac{G(x,y+h) - G(x,y)}{h} \, dx = \int_a^b \dfrac{\partial}{\partial y} G(x,y) \, dx \\ = \int_a^b \lim_{h \rightarrow 0} \frac{G(x,y+h) - G(x,y)}{h} \, dx .$$

Perhaps you want to show that

$$\tag{*}\frac{\partial^2}{\partial x \partial y}\int_a^x \left(\int_c^y f(u,v) \, dy \right) \, dx = f(x,y).$$

As you have written it $F(x,y)= \int_a^b \int_c^d f(u,v) \, du \, dv$ indicates that the function $F$ is a constant with zero partial derivatives since the integral on the RHS is a constant (real number) independent of $x$ and $y$.

Assuming that $f \in C(R)$ you can apply the fundamental theorem of calculus twice to prove (*).

First you must show that $G(u,y) = \int_c^y f(u,v) \, dv$ is continuous on $R$ and, consequently it follows, using a basic theorem for switching derivative and integral, that

$$\frac{\partial}{\partial y}F(x,y) = \frac{\partial}{\partial y}\int_a^xG(u,y) \, dy = \int_a^x\frac{\partial}{\partial y}G(u,y) \, du$$

An application of the fundamental theorem of calculus (FTC) yields

$$\frac{\partial}{\partial y}G(u,y) = \frac{\partial}{\partial y}\int_c^y f(u,v) \, dv= f(u,y).$$

Hence,

$$\frac{\partial}{\partial y}F(x,y) = \int_a^x f(u,y) \, du.$$

A second application of the FTC gives the desired result

$$\frac{\partial^2}{\partial x \partial y}F(x,y) = \frac{\partial}{\partial x }\int_a^x f(u,y) \, du = f(x,y).$$

You should attempt to finish yourself by applying Fubini's theorem to compute the mixed partial derivative with $x$ and $y$ reversed, and by providing justification for the first switching of derivative and integral.

• Do you think that my UPDATE is good justification for the first switching of derivate and integral? Commented Mar 4, 2017 at 19:44
• And if what I wrote is correct, applying Fubini's theorem I could justify the equality with the mixed partial derivates, because I could do something analogous taking the iteration of the integrals, right? Commented Mar 4, 2017 at 19:47
• On the first line you should write $\frac{\partial}{\partial y}F(x,y) = \frac{\partial}{\partial y}\int_a^xG(u,y) \, du$. You have $dy$ which is incorrect. Then you are correct up to the third equation. Your justication for switching the derivative and integral is OK. This would give $\frac{\partial}{\partial y}F(x,y) = \int_a^x f(u,y) \, du$.Then you can find $\frac{\partial}{\partial x}\frac{\partial}{\partial y}F(x,y) = f(x,y)$ just by fundamental theorem of calculus.
– RRL
Commented Mar 4, 2017 at 20:23
• You can use Fubini to switch the order of the integrals and follow the same steps to show $\frac{\partial}{\partial y}\frac{\partial}{\partial x}F(x,y) = f(x,y).$
– RRL
Commented Mar 4, 2017 at 20:25