# Thread: Prove two double integrals are equal

1. ## Prove two double integrals are equal

I need to prove:
$\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t) \frac{\partial^2\phi}{\partial x^2}(x, t) dx dt = \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t) \frac{\partial^2\phi}{\partial t^2}(x, t) dx dt$

Where $f(x - t)$, is a function of $x$ and $t$, and is merely piecewise continuous, so it doesn't have to be differentiable at certain points, and $\phi(x,t)$ is an arbitrary function also of $x$ and $t$, which is infinitely differentiable and has compact support.

My Attempt:

I tried to change variables by using $u = x - t$ and $v = x + t$, and I later realized that it basically became an exercise in trying to prove $\frac{\partial^2\phi}{\partial x^2} = \frac{\partial^2\phi}{\partial t^2}$ which didn't work out.

Can anyone give me a hint of some kind on how to try and proceed? Any suggestions would be greatly appreciated. Thanks in advance.

2. Originally Posted by mukmar
I need to prove:
$\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t) \frac{\partial^2\phi}{\partial x^2}(x, t) dx dt = \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t) \frac{\partial^2\phi}{\partial t^2}(x, t) dx dt$

Where $f(x - t)$, is a function of $x$ and $t$, and is merely piecewise continuous, so it doesn't have to be differentiable at certain points, and $\phi(x,t)$ is an arbitrary function also of $x$ and $t$, which is infinitely differentiable and has compact support.

My Attempt:

I tried to change variables by using $u = x - t$ and $v = x + t$, and I later realized that it basically became an exercise in trying to prove $\frac{\partial^2\phi}{\partial x^2} = \frac{\partial^2\phi}{\partial t^2}$ which didn't work out.

Can anyone give me a hint of some kind on how to try and proceed? Any suggestions would be greatly appreciated. Thanks in advance.
Suppose, to make life easier, that the function f is in fact differentiable. Then

\begin{aligned}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} &f(x - t) \frac{\partial^2\phi}{\partial x^2}(x, t) dx dt \\ &= \int_{-\infty}^{\infty}\Bigl\{\Bigl[f(x - t)\frac{\partial\phi}{\partial x}(x, t)\Bigr]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}f'(x-t)\frac{\partial\phi}{\partial x}(x, t)\,dx\Bigr\}dt \quad\text{\footnotesize (integration by parts)} \\ &= -\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f'(x - t)\frac{\partial\phi}{\partial x}(x, t)\,dxdt \quad\text{\footnotesize (because \tfrac{\partial\phi}{\partial x} vanishes at x=\pm\infty)} \\ &= -\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f'(x - t)\frac{\partial\phi}{\partial x}(x, t)\,dtdx \quad\text{\footnotesize (switching the order of integration)} \end{aligned}

. . . . . . . . .. $\displaystyle= \int_{-\infty}^{\infty}\Bigl\{\Bigl[f(x - t)\frac{\partial\phi}{\partial x}(x, t)\Bigr]_{-\infty}^{\infty} - \int_{-\infty}^{\infty}f(x-t)\frac{\partial^2\phi}{\partial x \partial t}(x, t)\,dt\Bigr\}dx$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . $\text{\footnotesize (integration by parts again, with respect to t this time)}$

. . . . . . . . .. $\displaystyle= -\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t)\frac{\partial^2\phi}{\partial x\partial t}(x, t)\,dtdx.$

That is symmetric in t and x, so it is equal to what you would get starting from the other end with $\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t) \frac{\partial^2\phi}{\partial t^2}(x, t) dx dt$ and going through the same procedure.

So the result is true for differentiable f. Then you'll have to think up some approximation argument for extending the result to piecewise continuous functions.

3. Thank you for the assisstance.

I have a question on what you said, what exactly do you mean by approximation argument?

4. Originally Posted by mukmar
I have a question on what you said, what exactly do you mean by approximation argument?
Let $\mathcal{F}$ denote the set of all functions f satisfying

$\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t) \frac{\partial^2\phi}{\partial x^2}(x, t) dx dt = \displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x - t) \frac{\partial^2\phi}{\partial t^2}(x, t) dx dt.$

We know that if f is differentiable then $f\in\mathcal{F}$. Suppose you can show that (1) $\mathcal{F}$ is closed under uniform limits, and (2) every continuous function is a uniform limit of differentiable functions. Then it will follow that every continuous function is in $\mathcal{F}$.

In fact, both (1) and (2) are true. (1) is a standard result about uniform convergence. For (2), you can uniformly approximate a continuous function by a sequence of differentiable functions $f_n$ if you define $f_n$ to be the average value of f over a short interval: $f_n(x) = \frac n2\displaystyle\int_{x-(1/n)}^{x+(1/n)}f(t)\,dt.$

To get the result for piecewise continuous functions, you would need to use a similar strategy. The set $\mathcal{F}$ is closed under bounded pointwise convergence (by the bounded convergence theorem), and every piecewise continuous function can be approximated (pointwise) by a bounded sequence of continuous functions.

5. So you're sure there's no "simple" way to prove this, i.e. just doing a change of variables in the original integrals somehow?

6. Originally Posted by mukmar
So you're sure there's no "simple" way to prove this, i.e. just doing a change of variables in the original integrals somehow?
No, I can't be sure that there is no simpler proof, but I don't see one.