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Prove that a differentiable function $f\colon I\to\mathbb{C}$ on an interval $I$ with bounded derivative is Lipschitz continuous, i.e. If $\lvert f'\rvert\leq L $ for some $L\in\mathbb{R}$, then for any $x_1,x_2\in I$ we have $$ \lvert f(x_1)-f(x_2)\rvert\leqslant L\lvert x_1-x_2\rvert. $$

Despite two little things, I think the proof should work as follows:

I think, we can choose some $c\in\mathbb{C}$ with $\lvert c\rvert =1$ such that $$ \lvert f(x_2)-f(x_1)\rvert = c\cdot (f(x_2)-f(x_1))~~(*) $$ since for $$ \frac{f(x_2)-f(x_1)}{\lvert f(x_2)-f(x_1)\rvert}=:v $$ we have $\lvert v\rvert =1$ and then we can define $c:=v^{-1}$.

Next, in order to have equation $(*)$, the LHS has to be the real part of the RHS, i.e. $$ \lvert f(x_2)-f(x_1)\rvert =\Re(c\cdot (f(x_2)-f(x_1)))=\varphi(x_2)-\varphi(x_1), $$ where $$ \varphi:=\Re(cf). $$

Now, since $f$ is differentiable on $I$, it is, in particular, continuous on $[x_1,x_2]$, hence $\varphi$ is also continuous on $[x_1,x_2]$.

I am not sure about the following question:

Do we also have that $\varphi$ is differentiable on $(x_1,x_2)$? (Q)

Assuming that we can answer question (Q) with YES, we could apply the mean value Theorem on $\varphi$, telling us that $$ \varphi(x_2)-\varphi(x_1)=(x_2-x_1)\varphi'(\xi) $$ for some $\xi\in (x_1,x_2)$.

By assumption, $\varphi'(\xi)=\Re(cf'(\xi))\leq\lvert cf'(\xi)\rvert\leq L$ and hence $$ \lvert f(x_2)-f(x_1)\rvert = \varphi(x_2)-\varphi(x_1)=(x_2-x_1)\varphi'(\xi)\leqslant L\lvert x_2-x_1\rvert. $$

I am also not completely sure if $$ \varphi' = \Re(cf') $$ is correct.

Despite the two things in the two yellow boxes, I am pretty sure the proof should work. It would be nice if you could give me some hints.

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  • $\begingroup$ You need to complex conjugate $c$. $\varphi$ is in general not differentiable if $0\in f((x_1, x_2))$. You could consider the squared absolute value. $\endgroup$
    – user251257
    Commented Jan 6, 2017 at 21:32
  • $\begingroup$ You mean $\lvert f(x_2)-f(x_1)\rvert = \bar{v} (f(x_2)-f(x_1))$? $\endgroup$
    – Rhjg
    Commented Jan 6, 2017 at 21:36
  • $\begingroup$ yes, or to circumvent the differentiability issue just set $c = \overline{f(x_2) - f(x_1)}$. $\endgroup$
    – user251257
    Commented Jan 6, 2017 at 21:38
  • $\begingroup$ Here is a related answer that does something similar math.stackexchange.com/a/2078872/27978. Essentially the idea is that if $\operatorname{re} (cz) \le K |c|$ for all $c$, then $|z| \le K$. In the above case, $K=L|x_1-x_2|$, $|c|=1$ and $z=f(x_1)-f(x_2)$. The answer to both of your yellow boxes is yes. $\endgroup$
    – copper.hat
    Commented Jan 6, 2017 at 21:48
  • $\begingroup$ @user251257 Sorry, but did not get why $\varphi$ isn't differentiable if $0\in f((x_1,x_2))$. Could you explain that, please? $\endgroup$
    – Rhjg
    Commented Jan 6, 2017 at 22:20

1 Answer 1

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Since $\operatorname{re}$ is continuous we have $\lim_{h \to 0} \operatorname{re} {cf(z+h)-cf(x) \over h} = \lim_{h \to 0} {\operatorname{re}(cf(z+h))-\operatorname{re}(cf(x)) \over h} = \operatorname{re} (cf'(x))$. That is, if $\phi(x) = \operatorname{re}(cf(x))$, then $\phi'(x) = \operatorname{re} (cf'(x))$.

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  • $\begingroup$ Why can we write the Re in front of the whole fraction, i.e. do not see the first equation. $\endgroup$
    – Rhjg
    Commented Jan 6, 2017 at 22:58
  • $\begingroup$ If $b$ is real, then $\operatorname{re} ({a \over b}) = {\operatorname{re} a \over b}$. $\endgroup$
    – copper.hat
    Commented Jan 6, 2017 at 23:31
  • $\begingroup$ Ah, of course!! h is real here... i thought it is complex but we are dealing with an interval. $\endgroup$
    – Rhjg
    Commented Jan 6, 2017 at 23:41

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