A problem on $C^k$ Can anyone help me solve the following problem ?
Let $a,k >0$, $a$ is real and  $k$ is integer. Consider the set $S$ of all function $f\in C^k([0,a])$ such that 
1) $f(0)=0$ and $f(a)=1$
2) $f^{(i)}(0)=f^{(i)}(a)=0$ for $i=1..k$
3) $0\leq f(x)\leq 1$ for all $x\in [0,a]$.
What is $\inf_{f\in S}\Vert f^{(k)}\Vert_{\infty}$ ?  
All I know is that the answer must be $a^kA$ where $A$  is the answer in the case $a=1$. I  am interested in the cases $k=3$ and $k=5$ in particular. I am also interested in the inverse problem: Given $K>0$ what is $a$ such that the anwser of the above problem is $K$. 
 A: If $f$ satisfies the assumptions, then so does $\tilde f(x)=1-f(a-x)$, and so does $\frac{1}{2}(f+\tilde f)$. The $L^{\infty}$ norm does not increase under this sort of symmetrization. So we may assume that $f(x)=1-f(a-x)$, in particular $f(a/2)=1/2$. 
Taylor's formula tells us that $f(a/2)=\frac{f^{(k)}(\xi)}{k!}(a/2)^{k}$. Hence $\|f^{(k)}\|_\infty\ge k!2^{k-1}/a^k$. To show this estimate is sharp, take $f(x)=x^k 2^{k-1}/a^k$ on $[0,a/2]$ and $f(x)=1-f(a-x)$ on $[a/2,a]$. 
If anyone objects that $f^{(k)}$  is not zero at $0$ and $1$, kick them in the shins. (And tweak $f$.)
(Added) The function constructed above has constant kth derivative on $[0,a/2]$. One way to tweak it is make the kth derivative constant on $[\epsilon,a/2]$ and linear on $[0,\epsilon]$, so that its value at $0$ is $0$. The constant value will have to be made a little bit bigger to compensate for the smaller derivative near zero. There is nothing conceptually hard here, but the computations may be tiresome. 
Second approach: Consider a slightly "compressed" function $f((1+2\epsilon)x)$, extended by $0$ to the left and by $1$ to the right. Shift it to the right a little bit, so that the function is constant in the neighborhood of either endpoint. Now the only problem is that it's not $C^k$ smooth. However, the standard mollification procedure will make it smooth. The supremum of $f^{(k)}$ will not increase under mollification, since the derivative of convolution is convolution with the derivative. And if the support of mollifier is small enough, the function will still be constant in the neighborhood of either endpoint. 
