A related Mean Value theorem result for complex functions Let $f:\mathbb{R}\rightarrow\mathbb{C}$ a differentiable function and I wonder if I can affirm that
$$\forall a,b\in \mathbb{R}\,\,\text{we have}\,\,\left|\frac{f(b)-f(a)}{b-a}\right|\leq |f'(c)|\,\,\text{for some $c\in(a,b)$}.$$
We know that is not possible extend the mean value theorem for $f:\mathbb{R}\rightarrow\mathbb{C}$ differentiable functions, that is
$$\forall a,b \in \mathbb{R}\,,\,\exists c\in(a,b)\,\,\,\text{with}\,f'(c)=\frac{f(b)-f(a)}{b-a}\in \mathbb{C} $$
is, in general, not true (see for example $f(t)= e^{ i t}$ in $(0,2\pi)$).
I've tried another way
$$\bigl|f(b)-f(a)\bigr|=\left| \int_a^b f'(\zeta)\ d\zeta\right| \leq  \int_a^b \bigl|f'(\zeta)\bigr|\ |d\zeta|\leq \sup_{\zeta\in [a,b]}\bigl|f'(\zeta)\bigr|\,|b-a|$$
Now, by Weierstrass theroem, exist $c_1,c_2\in[a,b]$ with $f'(c_1)\leq f'(x)\leq f'(c_2)\,\,\,\forall x\in[a,b]$ so that
$$\left|\frac{f(b)-f(a)}{b-a}\right|\leq |f'(c_2)|$$
What do you think about, is correct?
 A: There is an error towards the end.  When you apply the Weierstrass theorem, you are assuming that the derivative $f'$ is continuous, but there is no way to determine that from your hypotheses.  If you change your conditions to include that $f$ is continuously differentiable, then I believe you are correct.
EDIT
I found a proof of your result, which uses the mean value theorem.  
Regard $f$ as a vector-valued function $f:[a,b]\to\mathbb{R}^2$, and write $f(t)=(x(t),y(t))$.  Put $z=f(b)-f(a)$, and define $$\phi(t)=z\cdot f(t)$$ (here, $\cdot$ represents the inner product on $\mathbb{R}^2$).  Then $\phi$ is real valued and differentiable in $(a,b)$, so the mean value theorem furnishes a $c\in(a,b)$ such that 
$$ \phi(b)-\phi(a)=\phi'(c)(b-a)=(b-a)z\cdot f'(c). $$
Furthermore, observe that
$$ \phi(b)-\phi(a)= z\cdot f(b)-z\cdot f(a)=z\cdot z=\|z\|^2.$$
Combining our results, the Cauchy Schwarz inequality now gives
$$ \|z\|^2=(b-a)|z\cdot f'(c)|\leq(b-a)\|z\|\|f'(c)\|. $$
Switching back to complex notation, this gives us
$$ \left|\frac{f(b)-f(a)}{b-a}\right|\leq |f'(c)|. $$
Note:  I claim no originality.  This is Theorem 5.19 from Rudin's PMA.
