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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.

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  • $\begingroup$ What does $1$-Lipschitz mean? $\endgroup$ – zhw. May 15 at 17:31
  • $\begingroup$ @A 1-Lipschitz function means a lipschitz function with a lipschitz constant equal to $ 1 $. $\endgroup$ – MathOverview May 15 at 17:43
  • $\begingroup$ In that case I'd recommend trying Arzela-Ascoli. $\endgroup$ – zhw. May 15 at 17:44
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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 $h_0 \in H$.

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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

$$\int_0^1|f-g_n| <a +1/n,\,\,n=1,2,\dots$$

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.

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