For my current research, I need to prove the following:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_0^1 \frac{dC(q(x) + k'(q'(x) - q(x)))}{dk'}\,dk' = \int_0^1 \int_L^U p(q(x) + k(q'(x) - q(x)))(q'(x)-q(x)) dx dk[/tex]

where [tex]C(q(x)) = \int_0^1 \int_L^U p(kq(x)) q(x)\,dx\,dk [/tex]

Here's what I've tried using the definition of functional derivative:

[tex]

\frac{\partial C(q(x))}{\partial q(x)}

[/tex]

[tex]

= \lim_{\delta q(x) \rightarrow 0} \frac{C[q(x) + \delta q(x)] - C[q(x)]}{\delta q(x)}

[/tex]

[tex]

= \int_L^U \int_0^1 \frac{\partial p(kq(x))}{\partial q(x)}q(x) + p(kq(x)) dk dx

[/tex]

My guess is that

[tex]

\frac{dC(q(x) + k'(q'(x) - q(x)))}{dk'} = \frac{\partial C(q(x) + k'(q'(x) - q(x)))}{\partial (q(x) + k'(q'(x) - q(x)))} \frac{d(q(x) + k'(q'(x) - q(x)))}{dk'}

[/tex]

but I'm not sure what to do next. Any help will be greatly appreciated!

Thank you very much!

Alice

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# Functional derivative

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