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Let $B_r(x_0)$ be a ball in $\mathbb{R^l}$ and $f\in C^1(B_r(x_0)$,$\mathbb{R^m})$ such that $||Df(x)||_L\le M$ for all $x\in B_r(x_0)$. Then, I want to prove that for $x,y\in B_r(x_0)$,

$$|f(x) - f(y)| \le M|x-y|$$

So I want to use the function $F(t)=f(x+t(y-x))$ and the say that $F(1)-F(0)=\int_0^1F'(t)dt $ so the I have that

$$f(y)-f(x)=\int_0^{1} f'(x+t(y-x))(y-x)dt$$

and we have that $||f'(x)||_E<||f'(x)||_L$ because by definition we get $||Tx||_L=\inf\{ M :|Tx|\le M|x|\}$ so,

$$|f(y)-f(x)| \le (M)(r)|(y-x)|$$

But I think there is something wrong in the derivative.

Can someone help me to fix the problems of my proof, or provide another proof unsing the same idea please?

Thanks a lot in advance.

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  • $\begingroup$ Why the answer was deleted? $\endgroup$
    – user162343
    Oct 9, 2015 at 12:57
  • $\begingroup$ What are $L$ and $E$? $\endgroup$
    – robjohn
    Oct 9, 2015 at 15:23
  • $\begingroup$ Where? Well I dont have L or E :) $\endgroup$
    – user162343
    Oct 9, 2015 at 15:25
  • $\begingroup$ $||Df(x)||_L$ and $||f'(x)||_E$ $\endgroup$
    – robjohn
    Oct 9, 2015 at 16:42
  • $\begingroup$ ooo the norm in the space of linear transformations, and the euclidean norm :), Can you help me to recover the answer given by @JohnMa please?, or can you provide a better answer that mine please ajajja :)?, or what do you think about Christian's answer. $\endgroup$
    – user162343
    Oct 9, 2015 at 16:45

2 Answers 2

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By chain rule, we have

$$F'(t) = DF (x + t(y-x) ) \cdot (y-x).$$

Thus $$\|F'(t)\| = \|DF (x + t(y-x) ) \cdot (y-x)\| \le \|DF (x+ t(y-x))\|_L \ \|y-x\|$$

(see the remark) and

$$\begin{split} \|f(y)-f(x)\| &\le \int_0^1 \|F'(t)\| dt \\ & \le \int_0^1 \|DF (x+ t(y-x))\|_L \ \|y-x\| dt \\ &\le \int_0^1 M\|y-x\| dx \\ &= M\|y-x\|. \end{split}$$

Remark Note that as

$$\|DF(x)\|_L = \inf\{M : \|DF(x) (v)\|\le M \|v\|\ \forall v\}$$

we have

$$\|DF(x) (v)\|\le M \|v\|$$

for all $M > \|DF(x)\|_L$. Now for any $v\neq 0$,

$$\frac{\|DF(x) (v)\|}{\|v\|}\le M. $$

for all $M > \|DF(x)\|_L$. So we have

$$\frac{\|DF(x) (v)\|}{\|v\|}\le \|DF(x)\|_L. $$

(To see this, try to show that $>$ is impossible). Multiply $\|v\|$ on both sides gives

$$\|DF(x) (v)\|\le \|DF(x)\|_L \ \|v\|$$

for all $v\neq 0$. But the same equality holds also when $v = 0$. Thus it holds for all $v$. In particular for $v = y-x$, we have

$$\|DF(x) \cdot (y-x)\|\le \|DF(x)\|_L \ \|y-x\|.$$

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  • $\begingroup$ Ok but where do you the norm of the linear transformation? $\endgroup$
    – user162343
    Oct 9, 2015 at 0:56
  • $\begingroup$ Please see the edit @user162343 $\endgroup$
    – user99914
    Oct 9, 2015 at 1:02
  • $\begingroup$ Right :) I didn't understand the part of the arrow, and your noatation $DF(x)(v)$ Thnks in advance $\endgroup$
    – user162343
    Oct 9, 2015 at 1:07
  • $\begingroup$ So is the answer clear now? @user162343 $\endgroup$
    – user99914
    Oct 9, 2015 at 1:40
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    $\begingroup$ I've made an edit again. Now the dots now are matrix product (Note that $DF(x)$ is a matrix) @user162343 $\endgroup$
    – user99914
    Oct 9, 2015 at 1:50
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By the chain rule you have $$F'(t)=df\bigl(x+t(y-x)\bigr).(y-x)\qquad(0\leq t\leq1)\ ,$$ whereby $F'(t)$ is a vector. By definition of the norm $\|A\|$ of a linear map $A$ one has $|A.x|\leq\|A\|\>|x|$ for all vectors $x$. In the case at hand this implies that $$\bigl|df\bigl(x+t(y-x)\bigr).(y-x)\bigr|\leq \|df\bigl(x+t(y-x)\bigr)\|\>|y-x|\ .$$ Given your assumption on $df$ we therefore may write $$\bigl|F'(t)\bigr|\leq\|df\bigl(x+t(y-x)\bigr)\|\>|y-x|\leq M\>|y-x|\qquad(0\leq t\leq1)\ .$$ This implies $$\bigl|f(y)-f(x)\bigr|\leq\int_0^1\bigl|F'(t)\bigr|\>dt\leq M\>|y-x|\ .$$ There is not more to it.

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  • $\begingroup$ Right,and the linear transformation norm? $\endgroup$
    – user162343
    Oct 9, 2015 at 13:28
  • $\begingroup$ I am back @ChristianBlatter, so I didn't understand the part when you bound $|F'(t)|$, Thnks :) $\endgroup$
    – user162343
    Oct 9, 2015 at 16:50

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