Minimum norm problem over Lipschitz functions

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