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In Gelfand and Formin Calculus of Variation, the increment of the functional is

$$ \Delta J [h] = J[y+h] - J[y]$$

it is supposed that the increment can also be written in the form of

$$\Delta J[h] = \varphi [h] + \varepsilon ||h||$$

however I do not see how the 2nd definition of the functional increment picks up a $ \varepsilon ||h||$ term. What is the significance of the $\varepsilon ||h||$ term in the 2nd definition.

It is confusing what does the $\varepsilon$ multiply $||h||$ mean in the context of a functional

Hope I could get this confusion clarified, as it was not explained the reason why this supposition was made.

Arctic Char
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  • See Motivation for the differentiability of a function $f:D\subset\Bbb{R}^m\to\Bbb{R}$. There OP asks about finite-dimensional spaces, but the definition (and motivation and intuition) are virtually unchanged in the general Banach space setting. Or, see this for the general definition. Here is some extra motivation. Here is an example of using the definition to prove the chain rule. – peek-a-boo Jun 16 '22 at 03:07
  • As you'll hopefully recognize after reading those answers, the key concept of differential calculus is linear approximations: "actual change = linear part + 'small' error term". That's what the $\epsilon(h),|h|$ term represents. – peek-a-boo Jun 16 '22 at 03:12
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    I think the OP's source of confusion is notational. Gelfand & Formin (or rather Silverman) do not make the dependence of $\epsilon$ on $h$ clear in the text. – copper.hat Jun 16 '22 at 06:19
  • thanks @peek-a-boo I now understand that the $\varepsilon ||h||$ is the linear error term, but how does $\varepsilon$ multiplied with $||h||$ gives the dimension of an error so to speak ? – TheSprintingEngineer Jun 17 '22 at 15:17
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    It is because if you take the error $\epsilon(h) |h|$ and divide by $|h|$ (which is also what we do in single variable calculus; differential calculus is all avout first order approzimations) and take a limit, you get 0. This condition is giving a quantitative description of how small the error must be in order to say that a function is differentiable. – peek-a-boo Jun 17 '22 at 16:25

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