This problem seems like it should be quite easy/standard, but I've not found a solution written down anywhere.
Consider the set of 1-Lipschitz functions on the $[0,d]$ interval. Define the linear approximation, $$\hat{f} = \frac{d-x}{d}f(0) + \frac{x}{d} f(d),$$ for $x \in [0,d]$. Show that, $$\sup_{f\in\mathrm{Lip}_1}\Vert \hat{f} - f\Vert_\infty \leq C(d).$$
I believe that for this case, $C(d) = \frac{d}{2}$.
Solution attempts
- Using Poincaré inequalities, I think we can show an upper bound of $d$ (the diameter). We have that $\hat{f} - f$ is zero-trace, the diameter is $d$, but I struggled to get a tight enough bound on the gradient of $\hat{f} - f$.
- Jackson's Theorem (Corollary 7.5), seems to give a looser bound of $3d$. But this applies to arbitrary polynomial approximation and I expect is only tight asymptotically(?).
- I got $d/2$ via a fairly elementary construction. WLOG, consider $f$ such that $f(1) \geq f(0)$. Then the set of piecewise 1-Lipschitz linear functions is dense in 1-Lipschitz functions (generalizing Stone-Weierstrass). And the supremum is obtained via a PWL function with two segments, slope -1 and then slope +1 (basically just hand-waving, but correct?). This exactly defines the function $f$ in terms of it's endpoints, so we now have a maximization problem with respect to those endpoints. The maximum is achieved when the endpoints are equal, and the uniform norm is $d/2$ in this case. (Hiding some details here for brevity.)
