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Definition: Let $U\subseteq \Bbb R^m$ be an open set. Let $f: U \to \Bbb R^n$ be a function and $T: \Bbb R^m \to \Bbb R^n$ be a linear transformation. We say that $f$ is strongly differentiable at $x_0$, with derivative $T$, if for every $\epsilon>0$ there is $\delta >0$ s.t for any $x_1, x_2$ in $ \delta$-nbd of $x_0$ i.e $\|x_i-x_0\|<\delta$ for $i=1,2$ imply $\|f(x_2)-f(x_1)-T(x_2-x_1)\|<\epsilon\|x_2-x_1\|$.

Problem 1: Show that if $f$ is strongly differentiable at $x_0$ then it satisfies Lipschitz condition in a neighbourhood of $x_0$.

Problem 2 Show that if f is differentiable in a neighbourhood $U$ of $x_0$ & if $Df$ is continuous as a function of $x$ at $x_0$ then $ f$ is strongly differentiable at $x_0$ .

My attempt is:

1) $\|f(x_2)-f(x_1)-T(x_2-x_1)\|<\epsilon\|x_2-x_1\|\Rightarrow \|f(x_2)-f(x_1)\|<\epsilon\|x_2-x_1\| + \|T(x_2-x_1)\|$ .

Can anyone help me? What do I have to do next?

2) $||f(x_2)-f(x_1)-T(x_2-x_1)||<||f(x_2)-f(x_1)||+ ||T(x_2-x_1)||$ [because of definition of norm T, but why $||f(x_2)-f(x_1)||<\epsilon''||x_2-x_1||$]

Can anyone help me? What do I have to do next?

Ri-Li
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  • For 1) simply note that every linear map is Lipschitz (because we are in a finite dimensional vector space). One way to see this is to set $L = \max_{\Vert x\Vert = 1} \Vert Tx \Vert$ and derive $\Vert Tx - Ty \Vert = \Vert T(x-y) \Vert \leq L \Vert x-y \Vert$ by homogeneity. For 2): Do you know the multidimensional mean value theorem? – PhoemueX Oct 02 '14 at 09:46
  • Yeah I know. Can you write where & how to use it? – Ri-Li Oct 02 '14 at 11:03
  • Where did you find these problems ? – Tony Piccolo Oct 02 '14 at 14:42
  • Why Sir?? Is there any problem in it?? – Ri-Li Oct 04 '14 at 06:34
  • Note that Problem 2 has already been asked at https://math.stackexchange.com/questions/950247/. (Perhaps this question should be edited to remove Problem 2 and then retitled to briefly state Problem 1.) – Toby Bartels Oct 22 '17 at 04:20

1 Answers1

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I will answer only Problem 1, since Problem 2 was separately asked at Show that a function is strongly differentiable if it is continuously differentiable..

As PhoemeuX hinted, you should use the norm of $T$ (which is often written $\lVert{T}\rVert$ but which PhoemeuX wrote $L$); this exists because $\{x \mid \lVert{x}\rVert \leq 1\}$ is compact (and nonempty) and $T$ is continuous, which in turn are because $\mathbb{R}^m$ has finite dimension. (And in fact, when people work in infinite-dimensional Banach spaces, they explicitly require $\lVert{T}\rVert$ to exist (as a finite supremum if not a maximum) when defining a strongly differentiable function between such spaces, and then the Lipschitz condition still follows.)

To make this answer complete, I'll copy the definition of $\lVert{T}\rVert$ from PhoemeuX's comment (with a tiny change from $\|x\| = 1$ to $\|x\| \leq 1$ to automatically handle the case $m = 0$): $\lVert{T}\rVert = \max_{\lVert{x}\rVert \leq 1} \lVert{Tx}\rVert$. Then $\lVert{Ty}\rVert = \lVert{y}\rVert\,\lVert{T(y/\lVert{y}\rVert)}\rVert \leq \lVert{T}\rVert\,\lVert{y}\rVert$ (unless $y = 0$, when $\lVert{Ty}\rVert \leq \lVert{T}\rVert\,\lVert{y}\rVert$ still holds).

Then your argument can be completed: $$\lVert{f(x_2) − f(x_1)}\rVert < \epsilon\,\lVert{x_2−x_1}\rVert + \lVert{T(x_2−x_1)}\rVert \leq \epsilon\,\lVert{x_2−x_1}\rVert + \lVert{T}\rVert\,\lVert{x_2−x_1}\rVert \leq (\epsilon+\lVert{T}\rVert)\,\lVert{x_2−x_1}\rVert$$ so pick your favorite $\epsilon > 0$, get the corresponding $\delta$-neighbourhood, and use $\epsilon+\lVert{T}\rVert$ as the Lipschitz constant. (Thus you can make the Lipschitz constant as close to $\lVert{T}\rVert$ as you like by using a sufficiently small neighbourhood. To see that you cannot always make it $\lVert{T}\rVert$ exactly, use $m,n = 1$, $f(x) = x^2$, $x_0 = 0$.)

Toby Bartels
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