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Let $X$ and $Y$ be Banach spaces, $f:X\to Y$ and $\bar{x}\in X.$

We say that $f$ is strictly differentiable at $\bar{x}$ if there exists $\nabla f(\bar{x})\in L(X,Y)$ such that $$\lim_{x\to \bar{x}, u \to \bar{x},x\ne u}\frac{f(x)-f(u)-\nabla f(\bar{x})(x-u) }{\|x-u\|}=0.$$

We say that $f$ is continuously differentiable around $\bar{x}$ if there exists a neighborhood $U$ of $\bar{x}$ such that $f$ is Frechet differentiable at each point in $U$ and the mapping $\nabla f(\cdot):U \to L(X,Y)$ is norm to norm continuous.

The book I am reading states that if $f$ is continuously differentiable then it is also strictly differentiable at $\bar{x},$ but provides no proof. I have tried to prove this using standard $\epsilon-\delta$ arguments, but have failed. Can you provide this proof or at least point to a reference?

David C. Ullrich
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John D
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  • Hmm, Why does that calculation make sense? ie, $X$ and $Y$ maybe are very "different" – Martín Vacas Vignolo Dec 29 '17 at 13:31
  • To clarify, why could you divide an element of Y with a real number? – Martín Vacas Vignolo Dec 29 '17 at 13:32
  • ??? I don't see what needs to be proved - your definition of "strictly differentiable at $x$" appears to be identical to the definition of "Frechet differentiable at $x$... – David C. Ullrich Dec 29 '17 at 13:33
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    @MartínVacasVignolo ??? Do you know what the phrase "Banach space" means? – David C. Ullrich Dec 29 '17 at 13:33
  • First, this is a stronger notion of differentiability since you are varying the base point, but the proof that continuous diff implies strict diff is very easy and you should manage to do it by yourself. As a hint try to add and subtract terms in such a way you get the differential at $\overline{x}$ and some terms which are the difference of gradients, which by continuity is small. – Diesirae92 Dec 29 '17 at 13:41
  • @DavidC.Ullrich Note that it is not exactly Frechet differentiability since you have $f(x)-f(u)...$ instead of $f(x)-f(\bar{x}) $ and dividing by $|x-u|$ instead of $|x-\bar{x}|.$ Obviously, strict differentiability implies Frechet differentiability, but the converse do not hold – John D Dec 29 '17 at 13:45
  • @Diesirae92 Thanks for your comment. However, as I said, this is precisely what I tried, but I reach some limit from which I can not go any further. I think we may have to use some mean value theorem there. It doesn't seem to be trivial – John D Dec 29 '17 at 13:46
  • @Magnusseen Ah, right. Sorry, missed the bar on the $\overline $x$. – David C. Ullrich Dec 29 '17 at 13:50
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    @Diesirae92 Note that when playing with adding and subtracting terms the problem of the denominator($|x-u|$) will remain. That is the difficulty you will encounter. – John D Dec 29 '17 at 13:54

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Hint: Given $x$ and $u$, define $\phi:[0,1]\to X$ by $$\phi(t)=f(u+t(x-u)).$$Use the fact that $f$ is continuously differentiable to show that $\phi$ is continuously differentiable. In doing so you'll come up with an expression for $\phi'$ in terms of $\nabla f$.

Now take $$f(x)-f(u)=\int_0^1\phi'(t)\,dt$$and rewrite the integral in terms of $\nabla f$...

Ok, it goes like so: It follows from the definitions that $$\phi'(t)= \nabla f(u+t(x-u))(x-u).$$Hence $$f(x)-f(u)-\nabla f(\overline x)(x-u) =\int_0^1(\nabla f(u+t(x-u))-\nabla f(\overline x))(x-u)\,dt,$$ so$$\begin{aligned}||f(x)-f(u)-\nabla f(\overline x)(x-u)|| &\le\int_0^1||(\nabla f(u+t(x-u))-\nabla f(\overline x))(x-u)||\,dt \\&\le||x-u||\int_0^1||\nabla f(u+t(x-u))-\nabla f(\overline x)||\,dt \\&\le\epsilon||x-u||\end{aligned}$$if $||x-\overline x||<\delta$ and $||u-\overline x||<\delta$ (noting that it follows that $||u+t(x-u)-\overline x||<\delta$).

David C. Ullrich
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  • Thanks, @David C. Ulrich! After your hint, I completed the proof. I appreciate that you took the time to write it completely. – John D Dec 29 '17 at 15:31
  • @Magnusseen Yeah, from the comments you made it sounded like the hint would be enough for you - decided to add the details for other readers. Here's a question for you if you're bored: If $f:\mathbb R\to\mathbb R$ it's not hard to show that $f$ is strictly differentiable at $a$ if and only if $f$ is differentiable near $a$ and $\lim_{x\to a} f'(x)=f(a)$. Does the same hold for $f:X\to Y$? Of course one direction is proved by the argument above - I suspect that the answer is yes for the other directiion, haven't worked out the details. – David C. Ullrich Dec 29 '17 at 15:44
  • @DavidC.Ullrich: Your last comment seems not to be true. At https://math.stackexchange.com/a/210963/58577 you find a function which is strictly differentiable in $0$ but has points of non-differentiability approaching $0$. – gerw Jan 14 '19 at 12:06
  • @gerw Of course you're right - II don't know what I was thinking... – David C. Ullrich Jan 14 '19 at 14:10