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Define a functional $J$ as $J[f] = f'(x_0)$ where $f: \mathbb{R} \rightarrow \mathbb{R}$, $f'=\frac{\partial f}{ \partial x }$ and $x_0$ is a particular fixed point in $\mathbb{R}$. (i.e. given any function, $J[f]$ functional evaluates $f'$ at a fixed point $x_0$). We can write $J[f]$ as

$$ J[f] = \int_{x_1}^{x_2} \delta(x-x_0) f'(x) dx $$ where $x_0 \in [x_1, x_2]$ and $\delta$ is dirac delta function.

Question: What will be functional derivative of $J[f]$ with respect to $f$? $$ \frac{ \delta J }{ \delta f } = \; ? $$ Here, $\delta$ is functional derivative.

To find functional derivative, we can start with following.

\begin{align} \int \frac{ \delta J }{ \delta f } (x) \phi(x)dx &= \left[ \frac{d}{d \epsilon } J[f + \epsilon \phi] \right]_{\epsilon=0} \\ &= \left[ \frac{d}{d \epsilon } \int \delta(x - x_0)(f'(x) + \epsilon\phi '(x))dx \right]_{\epsilon=0} \\ &= \int \delta(x - x_0)\phi '(x)dx \end{align}

Problem: How do we proceed further? I looked into some of the examples of functional derivative (one can check wikipedia page of functional derivative for examples). They convert the term in the integral of right side of above equation in 2 parts. One part is $\phi(x)$ and other part is independent of $\phi$ which is equal to $\frac{ \delta J }{ \delta f }$ (by comparison with left side) but in this case, adapting similar approach is giving me $\frac{ \delta J }{ \delta f }$ which is dependent on $\phi$.

Any ideas are welcome. Thank you in advance.

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  • $\begingroup$ Integrate by parts your last expression to find that $$\int \frac{\delta J}{\delta f} \phi(x)\, dx = \int -\delta'(x-x_0) \phi(x)\,dx.$$ $\endgroup$ – Zachary Mar 27 at 23:23
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In distribution theory the functional derivative of OP's functional $J[f]:=f^{\prime}(x_0)$ is minus the derivative of the Dirac delta distribution:$$\frac{\delta J[f]}{\delta f(x)}~=~-\delta^{\prime}(x-x_0). $$

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