Minimum norm problem over Lipschitz functions I need to solve the following optimization problem in Luenberger's Optimization by Vector Space Methods. I believe we can find the dual but I'm having trouble here.
\begin{align}
\mathrm{minimize}\quad&||x||_2 \\
\mathrm{subject\:to}\quad&||x||_{\infty}\geq\epsilon \\
&x\in\mathrm{Lip}_L[0,1],
\end{align}
where $\mathrm{Lip}_L[0,1]$ is the space of Lipschitz continuous functions with constant $L<\infty$.
 A: First, notice that you can assume the first constraint to be an equality. That's because any sequence $f_n$ that approaches the infimum can be rescaled to meet the equality.
So now we have to solve $$\begin{align}
\mathrm{minimize}\quad&||x||_2 \\
\mathrm{subject\:to}\quad&||x||_{\infty}=\epsilon \\
&x\in\mathrm{Lip}_L[0,1],
\end{align}$$
Let $f\in\mathrm{Lip}_L[0,1]$ such that $\|f\|_\infty=\varepsilon$. Without loss of generality, we can assume that $\|f\|_\infty$ is the maximum of $f$ (just multiply $f$ by $-1$ if it were the minimum). Remember that $f$ is continuous and achieves that maximum, let's assume in $a\in[0,1]$:
$$f(a) = \|f\|_\infty=\varepsilon$$
Let's construct $f^*\in\mathrm{Lip}_L[0,1]$ as being a triangle function with slope $L$ and that has its maximum in $a$ with value $\varepsilon$:
$$f^*(x)=\left\{
\begin{split}
\varepsilon+(x-a)L & \,\,\,\,\,\text{if }\max\left(0,a-\frac \varepsilon L\right)\leq x \leq a \\
\varepsilon+(a-x)L & \,\,\,\,\,\text{if }a\leq x \leq \min\left(1,a+\frac \varepsilon L \right)\\
0 & \,\,\,\,\,\text{ otherwise.}
\end{split}\right.$$

We will prove that for all functions $f\in\mathrm{Lip}_L[0,1]$ such that $f(a)=\|f\|_\infty=\varepsilon$, we have
$$\|f^*\|_2\leq \|f\|_2$$
That will show that the solution of the problem is among $f^*$ and all its translated versions.
Onto the proof itself: For $|x-a| \leq \frac \varepsilon L$, it's easy to verify that
$$f(x) \geq f(a) - |f(x)-f(a)|\geq \varepsilon-L|a-x|=f^*(x)$$
Thus $$\int_0^1|f(x)^2dx \geq \int_{|x-a|\leq \frac \varepsilon L}|f(x)|^2dx  \geq\int_{|x-a|\leq \frac \varepsilon L}|f^*(x)|^2dx  =\int_0^1|f^*(x)|^2 dx$$
Therefore the solution is to be found among $f^*$ and its shifted versions. And there are two special values where the quadratic norm is minimal, it's when $a=0$ and $a=1$, the half triangles. And in those cases $$\|f^*\|_2 = \sqrt{\frac{\varepsilon^3}{3L}}$$
