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I'm self studying generalized functions, and it's been the third time or so that I come across the following statement :

If $\phi(x)$ is differentiable at $0$, then there exists a function $\psi(x)$ continuous at $0$ such that $\phi(x) = \phi(0) + x\psi(x)$.

Which theorem or result is used here? Is there a higher order version (i.e. in the case where $\phi$ is twice differentiable or more).

At first I used to tell myself it's kind of like a taylor expansion, say :

$$\phi(x) = \phi(0)+x\phi'(0) + x^2\epsilon(x)$$

but then $\phi'(0)+x\epsilon(x)$ is not guaranteed to be defined at $0$; well, if it were, then we'd have the continuity at $0$ due to $\lim_{x \to 0} \epsilon(x) = 0$.

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the_firehawk
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  • By generalized functions do you mean distributions? – N. S. Jan 27 '19 at 22:58
  • @N.S. yes. the book I use is called 'Generalized Functions : Theory and Applications' – the_firehawk Jan 27 '19 at 22:59
  • $\phi$ bounded (continuous) means $\psi(x)=\frac{\phi(x)-\phi(0)}{x}$ is bounded (continuous) for $x \ne 0$ and $\phi$ differentiable at $0$ means $\psi(x)$ is continuous at $0$. – reuns Jan 27 '19 at 23:00
  • I believe that's Caratheodory's formulation of differentiability( at 0). From my undergrad classes, I know its equivalent to being differentiable (at 0) but I can't find a source. There is some discussion here https://www.reddit.com/r/math/comments/60c9z7/why_isnt_the_caratheodory_derivative_more/?st=jrfiizeq&sh=4c170ece – Calvin Khor Jan 27 '19 at 23:07
  • This has further details and references as well as sketches of e.g. how this is equivalent to a Frechet derivative in Banach spaces

    https://math.stackexchange.com/questions/2321000/a-new-definition-of-derivative

     – Calvin Khor Jan 28 '19 at 11:43

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Hint

$$\phi(x) = \phi(0) + x\psi(x) \Leftrightarrow \\ \psi(x)= \frac{\phi(x)-\phi(0)}{x-0} \forall x \neq 0$$

$\phi$ is differentiable at $x=0$ if and only if $$\lim_{x \to 0} \psi(x)= \phi'(0)$$

In order to make $\psi$ continuous at $x=0$ you need to define $\psi(0)$ such that $$\lim_{x \to 0} \psi(x)= \psi(0)$$

SOOO You know what $\psi(x)$ should be for $x \neq 0$ and what $\psi(0)$ should be.

N. S.
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  • oh I see, extension by continuity. so we could've also defined $\psi$ for non-zero $x$'s the way I thought in the 1st place using taylor expansion and at $0$ we'd just put the limit of that same function and then we'd get the same $\phi'(0)$. thanks. – the_firehawk Jan 27 '19 at 23:06