Integration by parts for functions not n times continuously differentibal?

Question

Suppose $g:\mathbb{R}^2\rightarrow \mathbb{R}$ is a Lebesgue absolutely integrable function, define another function $f:\mathbb{R}^2\rightarrow \mathbb{R}$ as $f(y_1,y_2)=\int_{-\infty}^{y_1}\int_{-\infty}^{y_2}g(x_1,x_2)dx_1dx_2$. Another function $Q:\mathbb{R}^2\rightarrow \mathbb{R}$ is an infinitely differentiable and compactly supported function(a test function in $\mathbb{R}^2$). By Fubini's theorem, $f$ is well-defined, and it is continuous. I want to prove that $$\int_\mathbb{R}\int_\mathbb{R}f(y_1,y_2)(\partial^2 Q/\partial y_1\partial y_2)dy_1dy_2=\int_R\int_Rg(y_1,y_2)Q(y_1,y_2)dy_1dy_2.$$ Indeed this is a simple step in Rudin's functional analysis Page 168 and Rudin just commented here "by integration by parts". I wonder why Rudin could use integration by parts here for $f$ not twice continuously differentible. Indeed I just know the following relevant result for integration by parts(in Terence Tao's Intro to measure theory): a function $h:[a,b]\rightarrow R$ is absolutely continuous, and a function $m:[a,b]\rightarrow R$ is differentiable and its first derivative is continuous(i.e., once continuously differentiable) with $m$ being compactly supported in $[a,b]$, then the second fundamental theorem of calculus implies $$\int_a^bhm'=-\int_a^bh'm$$. I have no idea to apply this result to the previous problem and how Rudin proves it just by using integration by parts.

Answer 1

Fix $M,N>0$ and let $F(y_{1},y_{2})=\displaystyle\int_{-M}^{y_{2}}\int_{-N}^{y_{1}}g(x_{1},x_{2})dx_{1}dx_{2}$. Also fix $c<d$, $a<b$. Now we have \begin{align*} &\int_{a}^{b}\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})\int_{-M}^{y_{1}}\int_{-N}^{y_{2}}g(x_{1},x_{2})dx_{2}dx_{1}dy_{2}\\ &=\int_{a}^{b}F(y_{1},y_{2})\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})dy_{2}\\ &=F(y_{1},b)\dfrac{\partial Q}{\partial y_{1}}(y_{1},b)-F(y_{1},a)\dfrac{\partial Q}{\partial y_{1}}(y_{1},a)-\int_{a}^{b}\int_{-N}^{y_{1}}g(x_{1},y_{2})dx_{1}\dfrac{\partial Q}{\partial y_{1}}(y_{1},y_{2})dy_{2}, \end{align*} where we can choose $a<b$ such that the boundary vanishes, so \begin{align*} &\int_{a}^{b}\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})\int_{-M}^{y_{1}}\int_{-N}^{y_{2}}g(x_{1},x_{2})dx_{2}dx_{1}dy_{2}\\ &=-\int_{a}^{b}\int_{-N}^{y_{1}}g(x_{1},y_{2})dx_{1}\dfrac{\partial Q}{\partial y_{1}}(y_{1},y_{2})dy_{2}. \end{align*} Now we let $G(y_{1})=\displaystyle\int_{-N}^{y_{1}}g(x_{1},y_{2})dx_{1}$ (omitting the variable $y_{2}$), then \begin{align*} &\int_{c}^{d}\int_{a}^{b}\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})\int_{-M}^{y_{1}}\int_{-N}^{y_{2}}g(x_{1},x_{2})dx_{2}dx_{1}dy_{2}dy_{1}\\ &=-\int_{c}^{d}\int_{a}^{b}\int_{-N}^{y_{1}}g(x_{1},y_{2})dx_{1}\dfrac{\partial Q}{\partial y_{1}}(y_{1},y_{2})dy_{2}dy_{1}\\ &=-\int_{c}^{d}\int_{a}^{b}G(y_{1})\dfrac{\partial Q}{\partial y_{1}}(y_{1},y_{2})dy_{2}dy_{1}\\ &=-\int_{a}^{b}\int_{c}^{d}G(y_{1})\dfrac{\partial Q}{\partial y_{1}}(y_{1},y_{2})dy_{1}dy_{2}\\ &=-\int_{a}^{b}\left[\left(G(d)\dfrac{\partial Q}{\partial y_{1}}(d,y_{2})-G(c)\dfrac{\partial Q}{\partial y_{1}}(c,y_{2})\right)-\int_{c}^{d}g(y_{1},y_{2})Q(y_{1},y_{2})dy_{2}\right]dy_{1}, \end{align*} once again we choose $c<d$ such that the boundary vanishes, so \begin{align*} &\int_{c}^{d}\int_{a}^{b}\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})\int_{-M}^{y_{1}}\int_{-N}^{y_{2}}g(x_{1},x_{2})dx_{2}dx_{1}dy_{2}dy_{1}\\ &=\int_{a}^{b}\int_{c}^{d}g(y_{1},y_{2})Q(y_{1},y_{2})dy_{2}dy_{1}. \end{align*} Now use Lebesgue Dominated Convergence Theorem to deduce that whenever $M,N\rightarrow\infty$, we have \begin{align*} &\int_{c}^{d}\int_{a}^{b}\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})\int_{-\infty}^{y_{1}}\int_{-\infty}^{y_{2}}g(x_{1},x_{2})dx_{2}dx_{1}dy_{2}dy_{1}\\ &=\int_{a}^{b}\int_{c}^{d}g(y_{1},y_{2})Q(y_{1},y_{2})dy_{2}dy_{1}. \end{align*} Finally we note that \begin{align*} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})\int_{-\infty}^{y_{1}}\int_{-\infty}^{y_{2}}g(x_{1},x_{2})dx_{2}dx_{1}dy_{2}dy_{1}\\ &=\int_{c}^{d}\int_{a}^{b}\dfrac{\partial^{2}Q}{\partial y_{1}\partial y_{2}}(y_{1},y_{2})\int_{-\infty}^{y_{1}}\int_{-\infty}^{y_{2}}g(x_{1},x_{2})dx_{2}dx_{1}dy_{2}dy_{1}\\ &=\int_{a}^{b}\int_{c}^{d}g(y_{1},y_{2})Q(y_{1},y_{2})dy_{2}dy_{1}\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}g(y_{1},y_{2})Q(y_{1},y_{2})dy_{2}dy_{1} \end{align*} for some $c<d,a<b$ by the assumption that $Q$ being compactly supported.

Answer 2

The derivative of $f$ that is being used is $\frac{\mathrm{d}}{\mathrm{d} y_2}\frac{\mathrm{d}}{\mathrm{d} y_1}f(y_1, y_2)$ (note the order), which we already know is $g(y_1, y_2)$, by the definition of $f$ and the (first part of the) fundamental theorem of calculus. That is, the only twice differentiability in use is "once in each variable".

Answer 3

Integration by parts in the context of definite integrals is nicely understood if you simply forget about differentiation. Indeed, it is just a classical rendition of the more general statement Fubini's theorem.

To see why, let us examine the classical case. Assume that $f$, $g$ are Lebesgue integrable on $[a, b]$. If we write $F(x) = \int_{a}^{x} f(t) \, dt$ and $G(x) = \int_{a}^{x} g(t) \, dt$, then the integration by parts is proved in the following way:

\begin{align*} \int_{a}^{b} F(x)g(x) \, dx &= \int_{a}^{b} \left( \int_{a}^{x} f(t) \, dt \right) g(x) \, dx \\ &= \int_{a}^{b} \int_{a}^{b} f(t)g(x)\mathbf{1}_{\{t < x \}} \, dtdx \\ &= \int_{a}^{b} \int_{a}^{b} f(t)g(x)\mathbf{1}_{\{t < x \}} \, dxdt \tag{Fubini} \\ &= \int_{a}^{b} f(t)(G(b) - G(t)) \, dt \\ &= F(b)G(b) - \int_{a}^{b} f(t) G(t) \, dt. \end{align*}

For general constants of integration, it is straightforward to check that replacing $F$ and $G$ by $F + C_1$ and $G + C_2$, respectively, causes no harm to this formula. Combining this with the fundamental theorem of calculus for absolutely continuous functions, we obtain

Proposition. If $f$ and $g$ are absolutely continuous functions on $[a, b]$, then $$ \int_{a}^{b} f(x) g'(x) \, dx = [f(b)g(b) - f(a)g(a)] - \int_{a}^{b} f'(x)g(x) \, dx. $$

For your case, you can either utilize the above statement, or simply adopt the proof technique directly to establish the desired claim.