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Having not received an answer to my previous question, I'm trying to get the result using distributions theory. I've found the result for fixed integration limits here in theorem 4. Now $$\frac{d}{dx}\int_{a(x)}^{b(x)}f(x,\omega)d\omega=\frac{d}{dx}\int_{\mathbb{R}}\mathbf{1}_{[a(x),b(x)]}(\omega)\ f(x,\omega)d\omega=\\ \frac{d}{dx}\int_{\mathbb{R}}(H(\omega-a(x))-H(\omega-b(x)))\ f(x,\omega)d\omega=\\ \int_{\mathbb{R}}\left.\left.\frac{\partial}{\partial x}\right[(H(\omega-a(x))-H(\omega-b(x)))\ f(x,\omega)\right]d\omega=\\ \int_{\mathbb{R}}\left[(-\delta(\omega-a(x))a'(x)+\delta(\omega-b(x))b'(x))f(x,\omega)+ \mathbf{1}_{[a(x),b(x)]}(\omega)\ \frac{\partial f}{\partial x}(x,\omega)\right] d\omega=\\ f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}\frac{\partial f}{\partial x}(x,\omega)d\omega.$$ Here $\mathbf{1}_{[a,b]}(\omega)$ is the indicator function of segment $[a,b]$ and $H(x)$ is the Heaviside step function. The doubtful trick here is the usage of the chain rule to calculate $\frac{\partial}{\partial x}(H(\omega-a(x))$. The question is, is such usage of the chain rule legal? Is there some chain rule theorem for distributions at all? And is this proof correct?

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Trying to do a proper calculation of $\frac{\partial}{\partial x}H(\omega-a(x))$ gives $$ \left< \frac{\partial}{\partial x}H(\omega-a(x)), \phi(\omega) \right> = \frac{\partial}{\partial x} \left< H(\omega-a(x)), \phi(\omega) \right> \\ = \{ \tilde\omega = \omega-a(x) \} \\ = \frac{\partial}{\partial x} \left< H(\tilde\omega), \phi(\tilde\omega+a(x)) \right> \\ = \frac{\partial}{\partial x} \int H(\tilde\omega) \, \phi(\tilde\omega+a(x)) \, d\tilde\omega \\ = \frac{\partial}{\partial x} \int_0^\infty \phi(\tilde\omega+a(x)) \, d\tilde\omega \\ = \int_0^\infty \frac{\partial}{\partial x} \phi(\tilde\omega+a(x)) \, d\tilde\omega \\ = \int_0^\infty a'(x) \, \phi'(\tilde\omega+a(x)) \, d\tilde\omega \\ = \int H(\tilde\omega) \, a'(x) \, \phi'(\tilde\omega+a(x)) \, d\tilde\omega \\ = a'(x) \, \int H(\tilde\omega) \, \phi'(\tilde\omega+a(x)) \, d\tilde\omega \\ = a'(x) \, \left< H(\tilde\omega), \phi'(\tilde\omega+a(x)) \right> \\ = - a'(x) \, \left< H'(\tilde\omega), \phi(\tilde\omega+a(x)) \right> \\ = - a'(x) \, \left< H'(\omega-a(x)), \phi(\omega) \right> $$ This shows that $\frac{\partial}{\partial x}H(\omega-a(x)) = -a'(x) \, H'(\omega-a(x)).$

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