Question. Consider a continuous function $ f: [0,1] \to \mathbb {R} $. Prove that there exists a function $ g: [0,1] \to \mathbb{R} $ which is 1-Lipschitz, satisfies $ g (0) = 0 $ and has the property of which:
$$
\int_{0}^{1}|f(t)-g(t)|dt =\inf\left\{\left.\int_{0}^{1}|f(t)-h(t)|dt \;\;\right| h:[0,1]\to \mathbb{R}\mbox{ is 1-Lipschitz } \mbox{ and } h(0)=0 \right\}
$$
A 1-Lipschitz function means a lipschitz function with a lipschitz constant equal to $ 1 $.
My attempt. I have noticed that if $ f (x) \geq x $ (respectively $ f (x) \leq -x $) the solution is trivially $ g (x) = x $ (respectively $ g (x) = – x $).
Although I have not been able to prove it, it seems intuitive to me that if $ f $ is a polynomial function then the graph of $ g $ will be a polygonal chain in $ [0,1] \times [0,1] $.
If in fact the graph of $ g $ is a polygonal chain in $ [0,1] \times [0,1] $ when $ f $ is a polynomial function, then I think we can apply to Stone-Weierstrass theorem to show that the $ g $ function of the above question exists for any function.
Best Answer
Let $H=\{ h : [0, 1] \rightarrow \mathbb{R} \ | \ h(0)=0 \ and \ \forall x,y \in [0, 1], |h(x)-h(y)| \le |x-y| \}$. Since $H$ is uniformly bounded by $1$ and equicontinuous, it follows, by Arzela-Ascoli, that $H$ is compact in the uniform convergence topology.
Define $\phi : H \rightarrow \mathbb{R}$ by $\phi(h)=\int_{0}^1 |f(t)-h(t)|dt$. Given $h, g \in H$, we have
$$ |\phi(h) - \phi(g)| = | \int_0^1 (|f(t) - h(t)| - |f(t)-g(t)|)dt | $$ $$ \le \int_0^1 | \ |f(t) - h(t)| - |f(t)-g(t)| \ |dt $$ $$ \le \int_0^1 |h(t) - g(t)|dt$$ $$ \le \sup_{t \in [0, 1]} |h(t)-g(t)|$$
So $\phi$ is continuous. Therefore it attains its minimum at a point $h_0 \in H$.