Functional derivative of normal function

[ChrisPhys]

I can't convince myself whether the following functional derivative is trivial or not:

$\frac \delta {\delta \psi(x)} \big[ \partial_x \psi(x)\big],$

where $\partial_x$ is a standard derivative with respect to $x$.

One could argue that

$\partial_x \psi(x) = \int dx' [\partial_{x'} \psi(x')] \delta (x - x') = - \int dx' \psi(x') \partial_{x'} \delta (x - x'),$

assuming there is no boundary term in integration by parts.

In this case, the functional derivative would give

$\frac \delta {\delta \psi(x)}\Big[ - \int dx' \psi(x') \partial_{x'} \delta (x - x') \Big] = - \partial_{x'} \delta (x - x')\Big|_{x'=x} = 0.$

Any thoughts? Is this rigorous?

 

[Orodruin]

You really should use different variables for the function you are differentiating and the function you are taking the derivative with respect to ...

What makes you think that the last expression evaluates to zero? In the distribution sense it is the derivative of the delta distribution.

 

[ChrisPhys]

To address your last point:

What makes you think that the last expression evaluates to zero? In the distribution sense it is the derivative of the delta distribution.

I just viewed the $\delta$ function as the limit of a Gaussian, whose derivative at zero is zero. Is that an error?

 

[Orodruin]

To address your last point:

I just viewed the $\delta$ function as the limit of a Gaussian, whose derivative at zero is zero. Is that an error?

This then falls back on your original problem with not using different variables for the function you are differentiating and the function you are differentiating with respect to. Your question should be: "What is $\delta (\partial_\mu \psi(x))/\delta \psi(y)$?"
The result should be a distribution which picks out the derivative of a function, i.e., the derivative (in the distributional sense) of the delta distribution.

 

[ChrisPhys]

@Orodruin
Thank you, that makes sense. I appreciate your help.

 

[Gracie thomas]

Hi
To informally guess the derivative, I use the 'little o' notation and then check the details. However, for the details. I think it is easiest to use composition rule, ie, if f,g are (Fréchet) differentiable with appropriate domains/ranges, then Df∘g(x)=Df(g(x))Dg(x).
Thanks.

 

[Orodruin]

Hi
To informally guess the derivative, I use the 'little o' notation and then check the details. However, for the details. I think it is easiest to use composition rule, ie, if f,g are (Fréchet) differentiable with appropriate domains/ranges, then Df∘g(x)=Df(g(x))Dg(x).
Thanks.

This thread is about functional derivatives, not derivatives of composite functions.