Statement
$\forall L \in \mathbb{R}^+$, and $\forall f:\mathbb{R}\to\mathbb{R}$ for which
- $f$ isn't differentiable at point $a\in\mathbb{R}$
- $f$ is infinitely differentiable everywhere else (on $\mathbb{R}\setminus\{a\}$)
- $\lim_{x\to a-} f'(x) \ne \lim_{x\to a+} f'(x)$
$\forall p:\mathbb{R}\to\mathbb{R}$ polynomials the following holds:
$$\Rightarrow |\{x\in\mathbb{R}: |f(x)-p(x)|>L\}| = |\mathbb{R}|$$
Explanation
This function $f$ has a nondifferentiable point $a$, but it is differentiable in an area around $a$, with the limits of the derivative of $f$ being different on both ends on $a$. For example the function
$$f(x)=|x|$$
has such a point at $a=0$, with the limits of its derivative being $-1$ from the left, and $+1$ from the right:
$$\lim_{x\to a-} f'(x) = \lim_{x\to a-} -1 = -1 \ne 1 = \lim_{x\to a+} 1 = \lim_{x\to a+} f'(x)$$
In general, these functions look something like this near the point $a$ (Here I set $a=0$):
The statement says that such functions cannot be approximated by any polynomial $p$, because no matter what polynomial we choose, there will always be uncountably infinitely many $(|\mathbb{R}|)$ points that are arbitrarily far away from $f$ (here, distance $L$ away).
Intuition
For $f(x) = |x|$, this seems to hold empirically. If we try to fit a Lagrange polynomial onto $n=3,5,7,9,11,13$ points located on the $|x|$ function, we get (an increasingly worse and worse) result:
While the polynomial seems to fit nicely near the origin, further away, it completely fails to capture the shape of $|x|$.
Similarly, for a generalized absolute value function
$$V_{l,r}(x) = \begin{cases}lx \quad \text{if } x < 0 \\rx \quad \text{if } x \ge 0 \end{cases}$$
which has a slope of $l$ on $\mathbb{R}^-$ and a slope of $r$ on $\mathbb{R}^+$, we can assume that the above method would also not work.
Potential proof
We would first of all need to rigorously prove that the statement holds for $V_{l,r}$ for any $l$ (left) input and any $r$ (right) input. This helps because any function defined in the statement is locally a generalized absolute value function $V_{l,r}$ near $a$.
So after seeing that polynomial approximation doesn't work for $V_{l,r}$, it also cannot work for functions defined in the statement, since it already doesn't work in the local area of $a$.
However, there is an issue. The fact that the approximation doesn't work locally, does not directly mean that the approximation polynomial will be distance $L$ away from $f$ in uncountably infinitely many points. As we saw earlier, we can give a good approximation for $f$ near $a$, but the approximation only fails further away from $a$.
Nonetheless, the fact that polynomial approxmation doesn't work for the "simplest case" of $|x|$, would likely mean that it also doesn't work for more complex functions which are also non-differentiable at a point.
Context
I have answered this question about a "No Free Lunch" theorem for polynomial approximation. This would be my proposed theorem, but I couldn't rigorously prove it myself. It intuitively seems to hold, as long as it is true that $f(x)=|x|$ cannot be efficiently approxmated by polynomials, which again, seems to hold from testing.