Extending a $C^{1}([0, 1/2])$ function to be $C^{1}([0, 1])$ while preserving upper and lower bounds Suppose $f \in C^{1}([0, 1/2])$ such that $1 \leq f(x) \leq 2$ for all $0 \leq x \leq 1/2$ and $f'(0) = 0$. Does there exist a $g \in C^{1}([0, 1])$ such that 


*

*$1 \leq g(x) \leq 2$ for all $0 \leq x \leq 1$

*$g'(0) = 0$

*$f(x) = g(x)$ for all $0 \leq x \leq 1/2$?


From the Tietze extension theorem there is a continuous $g$ such that $f = g$ on $[0, 1/2]$ and $1 \leq g \leq 2$ on $[0, 1]$. But this $g$ may not be differentiable and so this $g$ obeys the first and third conditions but not the second.
One can also define a candidate $g$ by
\begin{align}
g(x) = \begin{cases}
f(x) & \text{ if } x \in [0, 1/2]\\
f(1/2) + f'(1/2)(x - 1/2) & \text{ if } x \in [1/2, 1]
\end{cases}
\end{align}
and this obeys $g \in C^{1}([0, 1])$, $g'(0) = 0$ and $f = g$ on $[0, 1/2]$ but then one doesn't (necessarily) have $1 \leq g \leq 2$ on $[0, 1]$. So this candidate obeys the second and third conditions but not the first.
Finally, one can define $g(x) = f(x/2)$ which obeys the first two conditions but not the third.
 A: 
(Possibly, there are some typos in the QUESTION).


In general, there is no such $g$. Indeed, let
$$ \forall_{x\in\left[0;\frac 12\right]}\quad
      f(x)\ :=\ 1\,+\,4\cdot x^2 $$
Then $\ f'(0)=0,\ $ and
$$ \forall_{x\in\left[0;\frac 12\right]}\quad 1\le f(x)\ \le 2$$
However, $\ f'\left(\frac 12\right) = 4>0 $ hence every $\ C^1$-extension $g$ of $f$ over $\ [0;1]\ $ would have values $\ >\ 2\ $ in a small right neighborhood of $\ \frac 12.$
A: Not positive about this but here is my attempt. 
Let $f$ be a function satisfying all of the criteria you list and suppose that $f(1/2) = 2$ and $f'(1/2) > 0$. Now if $g$ is a continuous extension of $f$ on $[0,1]$. Then $g(1/2) = 2$ and $g'(1/2) = f'(1/2)>0$. These both follow from the condition that $f$ and $g$ are equal on $[0,1/2]$. Since $g$ is $C^1$, $g'$ is continuous on $[0,1]$ and hence there exists a neighborhood around $1/2$ where $g'$ is positive. This implies that in this neighborhood, $g$ is increasing, but $g$ cannot exceed $g(1/2) = 2$. Therefore the extension you want does not exist in general.
