Minimum gradient of curve required to bound fixed area. I'm trying to calculate
$$\inf\big\{\Vert f'\Vert_\infty : f\in C^1[0, 1], f \ge 0, f(0)=a, f(1)=b, \Vert f \Vert_1 =D\big\}$$
Intuition tells me that the infimum should be the gradient of a straight line segment to a point $(0.5, y)$ with $y$ chosen appropriately to bound the correct area. However I can't think how to go about proving this.
Also if I wanted to introduce a further constraint on higher order derivatives, how could I plug this in - say $\Vert f''\Vert_\infty < \lambda$  ?
 A: This is a (not very detailed in one part) partial solution for sufficiently large $D$, where the condition$f\ge 0$ is ignored and we later have see for which $D$ the minimizing $f$ actually fulfills this condition.
Similiar to what the OP suggested, we want to find a point $(x_0,y_0)$ with $0 < x_0 < 1$and consider the piecewise linear function 
$$f_0(x) = 
\left\{ 
        \begin{matrix}
        a+\frac{y_0-a}{x_0}x, &\text{for }0 \le x \le x_0 \\
        y_0+\frac{b-y_0}{1-x_0}(x-x_0), &\text{for }x_0 \le x \le 1. \\        
        \end{matrix} 
\right.$$
First we note that this is well defined, as both branches give $f_0(x_0)=y_0$, and in addition $f_0(0)=a, f_0(1)=b$ holds. 
This $f_0$ is usually not differentiable at $x_0$, but it should be clear that it can be approximated by a series $g_n(x)$ of such functions that only change $f_0$ in a smaller and smaller neighbourhood of $x_0$ and monotonously change the derivative from the initial $\frac{y_0-a}{x_0}$ to the later $\frac{b-y_0}{1-x_0}$. That means $\lim_{n\to \infty} \Vert g_n\Vert_1 = \Vert f_0\Vert_1$ is possible to achieve as well as $\lim_{n\to \infty} \Vert g'_n\Vert_\infty = \Vert f'_0\Vert_\infty$.
What we want to find is the point $(x_0,y_0)$ that satifies 2 condtions:
$$\Vert f_0\Vert_1 = D,$$
$$\frac{y_0-a}{x_0} = - \frac{b-y_0}{1-x_0}$$
The first should be clear, the second means that the 2 branches have the same absolute value in $f'_0$.
I'll skip the details of actually determining those values, the first equation leads to a linear equation in $x_0$ and $y_0$, the second to an equation containing the term $x_0y_0$ and otherwise linear terms. Geometric considerations lead to the fact that this has exactly one solution with $0 < x_0 < 1$ if $D \neq \frac{a+b}2$ (in that case every $(x_0,y_0)$ that lies on the line segment from $(0,a)$ to $(1,b)$ is a solution).
That means we have now found a function $f_e$ that equals $f_0$ for this specific choice of $(x_0,y_0)$, that fulfills the conditions and has a certain $\Vert f'_e\Vert_\infty$.
We now prove that $\Vert f'_e\Vert_\infty$ is the infimum under consideration. Depending on the sign of $D-\frac{a+b}2$ the proof will be slightly different, let's start with
$D \ge \frac{a+b}2$. Then because of the definition of $f_e$ and the condition $\Vert f_e\Vert_1=D$, we have $f_e$ being at or above the line $y=a+(b-a)x$ that connects $(0,a)$ and $(1,b).$ It also means the first branch of $f_e$ is increasing, while the second is decreasing.
Let $f$ be any arbitary function that fulfills the conditions to be considered for the infimum. If $f(x_1) > f_e(x_1)$ for any $0 < x_1 \le x_0$, then by the mean value theorem there is an $x_2 \in (0,x_1)$ with $f'(x_2)=\frac{f(x_1)-f(0)}{x_1} > \frac{f_e(x_1)-f_e(0)}{x_1} = \Vert f'_e\Vert_\infty$, which imples $\Vert f'\Vert_\infty > \Vert f'_e\Vert_\infty$.
Similiarly, if there was $f(x_1) > f_e(x_1)$ for any $x_0 \le x_1 < 1$, then there was an $x_2 \in (x_1,1)$ with $f'(x_2)=\frac{f(1)-f(x_1)}{1-x_1} < \frac{f_e(1)-f_e(x_1)}{1-x_1} = -\Vert f'_e\Vert_\infty$. Again (recalling that $f'(x_2)$ is negative), we see that $\Vert f'\Vert_\infty > \Vert f'_e\Vert_\infty$ would be implied.
Recap: We've shown that any function that fulfills the conditions to be considered for the infimum, if it wants a smaller $\Vert f'\Vert_\infty$ than our $f_e$, it has to be at or below $f_e$. Now if it is equal to $f_e$, then it can't have a smaller $\Vert f'\Vert_\infty$. If it is not equal, it must be smaller at some point $x_3$, and by continuity in some neighborhood of $x_3$. But that means $\Vert f\Vert_1 < D$, which is a contradiction.
The case $D \le \frac{a+b}2$ works basically the same, by now showing that any $f$ that goes below $f_e$ has a higher $\Vert f'\Vert_\infty$ than $f_e$, and then showing that the opposite case implies $\Vert f\Vert_1 > D$.
Now, in all of this the condition $f \ge 0$ was dropped. If our $f_e$ fullfills that condition, the stated infimum is obviously also the infimum with the condition added. This can be checked in a given cases by comparing $y_0$ with 0.
If not, then a more detailed analysis is needed. Basically if $D$ is to small, $f$ needs to go down to 0 fast, such that the integral doesn't go over $D$. In that case, a 3-piecewise function seems to be the most natural idea: From $(0,a)$ to $(x_1,0)$ to $(x_2,0)$ to $(1,b)$. Again, the derivatives in the first and last branch should have equal value and opposing signs.
