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In the book Calculus of variations by I. M. Gelfand, the differential of a functional is defined in the following way, and its uniqueness is proven:

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However, it seems strange to me that the expression $\Delta J[h]=\varphi[h]+\epsilon||h||$ shouldn't be written as $\Delta J[h]=\varphi[h]||h||+\epsilon||h||$ instead. As far as I understand, a similar expression holds for the derivative of a function of a single real variable:

$f(a+h)-f(a)=f'(a)h+\Psi(h)=f'(a)h+\Psi_1(h)h$

where $\Psi(h)$ is a function such that $lim_{h\to 0}\Psi_1(h)=0$. But here there is an $h$ multiplying $f'(a)$, why shouldn't there be a $||h||$ multiplying $\varphi[h]$ in the case of a functional?

And there is something else that doesn't make sense to me either in the proof of Theorem 1. Taking as hypothesis that

$\Delta J[h]=\varphi_1[h]+\epsilon_1||h||$ $\Delta J[h]=\varphi_2[h]+\epsilon_2||h||$

why can we conclude that $\varphi_1[h]-\varphi_2[h]=\epsilon_2||h||$? Wouldn't it be $\epsilon_1-\epsilon_2$ instead of $\epsilon_2$? Maybe there is a typo and they meant to define a new infinitesimal $\epsilon_0=\epsilon_1-\epsilon_2$, or am I interpreting this totally wrong?

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In the real variable case, the linear functional is $h \mapsto f'(a) h$. It is not $f'(a)$ itself. Accordingly, in the general case, the linear functional is $h \mapsto \varphi(h)$; $\varphi$ includes both the "$f'(a)$" and the "multiplication by $h$" from the real variable setting.

I agree with you that at the end there is a problem, that $\varepsilon_2$ should not be the same $\varepsilon_2$ from earlier (though this "re-use" of symbols for quantities that we care little about is common in functional analysis).

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  • $\begingroup$ Thank you! But the $\varphi[h]$ doesn't literally have $||h||$ as a factor in whatever its expression might be, right? Because then it wouldn't be a linear functional, as the norm of the sum does not equal the sum of the norms... It's just that it is defined that way, isn't it? $\endgroup$ – Wild Feather Feb 4 '18 at 18:15
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    $\begingroup$ @WildFeather That's right. You can relate this to the notion of directional derivative that you probably learned in multivariable calculus by noticing that for $h \neq 0$ you have $\varphi(h)=\varphi(h/\| h \|) \| h \|$. $\endgroup$ – Ian Feb 4 '18 at 18:17

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