Minimum norm problem over Lipschitz functions

functional-analysisoptimizationvector analysisvector-spaces

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 $$L<\infty$$.

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 $$\|f^*\|_2 = \sqrt{\frac{\varepsilon^3}{3L}}$$