# Prove that there exists a function $g: [0,1] \to \mathbb{R}$ which is 1-Lipschitz, satisfies $g (0) = 0$ and has the following property.

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.

• What does $1$-Lipschitz mean? – zhw. May 15 at 17:31
• @A 1-Lipschitz function means a lipschitz function with a lipschitz constant equal to $1$. – MathOverview May 15 at 17:43
• In that case I'd recommend trying Arzela-Ascoli. – zhw. May 15 at 17:44

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 $$h_0 \in H$$.

Let $$G$$ be the family of $$g$$'s you defined. Then $$G$$ is uniformly bounded and equicontinuous on $$[0,1].$$ Also let $$a$$ be the infimum you defined.

Now there exists a sequence $$g_n$$ in $$G$$ such that

$$\int_0^1|f-g_n|

By Arzela-Ascoli, $$g_n$$ has a subsequence $$g_{n_k}$$ that converges uniformly on $$[0,1]$$ to some $$g_0\in C[0,1].$$ It's easily vefied that actually $$g_0\in G.$$ We then see

$$a\le \int_0^1|f-g_0| = \lim_k \int_0^1|f-g_{n_k}| \le \lim_k (a+1/n_{n_k})=a.$$

Thus $$g_0$$ is the desired minimizer.