Given $\phi \in C^{1,b}(R)$, find $\phi_n$ countably piecewise affine functions whose derivatives converge to $\phi'$ uniformly where differentiable

Question

Let $\phi \in C^{1}(\mathbb R)$ with bounded derivative. I am trying to build $\phi_n$ a sequence of countably piecewise affine functions, s.t. $\phi_n'$ converges uniformly to $\phi'$ on $N^c$, where $N$ is the (countable) set of non-differentiability points of all $\phi_n$.

---

My first try is to consider a countably piecewise affine interpolant, i.e. defined $J_{i,n} = [i/n, (i+1)/n]$, let $$ \phi_n(t) = \phi\left(\dfrac{i}{n} \right) + n\left(\phi\left(\dfrac{i+1}{n}\right) - \phi\left(\dfrac{i}{n}\right)\right) \left(t- \dfrac{i}{n} \right) \quad \text{ if } t \in J_{i,n}, \forall i \in \mathbb Z. $$ Now, if $\phi \in C^2$ with $\phi'' \leq L$, I could estimate for $t \in J_{i,n}$ $$ \phi'_n(t) - \phi'(t) \leq \phi'(\xi) - \phi'(t) \leq L|J_{i,n}| ,$$ (for some $\xi \in J_{i,n}$), thus I would have uniform convergence as $n \to \infty$. But what if $\phi$ is just $C^1$ with bounded derivative?

Answer 1

You ask about $\phi_n\to\phi$, but I gather what you're actually concerned with is $\phi_n'\to\phi'$. This happens at every point of $N^c$, just from the definition of the derivative, more or less.

If $x\in N^c$ then for every $n$ there exists $i$ such that $$x\in(i/n,(i+1)/n)=(a_n, b_n).$$ Also $$\phi_n'(x)=\frac{\phi(b_n)-\phi(a_n)}{b_n-a_n}.$$ Hence $\phi_n'(x)\to\phi'(x)$, by the following elementary lemma:

Lemma Suppose $f$ is differentiable at $x$, $a_n<x<b_n$ and $b_n-a_n\to0$. Then $(f(b_n)-f(a_n))/(b_n-a_n)\to f'(x)$.

Hint Write $(f(b_n)-f(a_n))/(b_n-a_n)$ as a convex combination of $(f(b_n)-f(x))/(b_n-x)$ and $(f(x)-f(a_n))/(x-a_n)$, both of which tend to $f'(x)$ by definition.

EDIT: Now it turns out that we actually want $\phi_n'\to\phi'$ uniformly on $N^c$. I do not believe this is true for the $\phi_n'$ as above.

If $\phi_n$ is as above then the $\phi_n'$ do converge uniformly to $\phi'$ on bounded subsets of $N^c$. And if we assume that $\phi'$ is uniformly continuous then $\phi_n'\to\phi'$ uniformly on $N^c$. This follows from the lemma above plus

Lemmma 2 If $f'$ is uniformly continuous then $(f(x+h)-f(x))/h\to f'$ uniformly.

Hint Mean Value Theorem.

Now if $\phi'$ is just continuous and bounded (I don't see how boundedness helps) but not uniformly continuous one could do more or less the same thing. Except that instead of taking a uniform partition as above one would need to use a partition adapted to the modulus of continuity of $\phi$, with shorter and shorter intervals as you approach infinity, corresponding to the fact that $\phi'$ is less and less continuous.

Answer 2

Basically the same as David Ullrich's proof, but maybe a little shorter: Suppose $f\in C^1 [a,b].$ Let $\epsilon>0.$ By the uniform continuity of $f'$ on $[a,b],$ there exists $\delta > 0$ such that $|y-x|<\delta \implies |f'(y)-f'(x)| < \epsilon.$ Let $P = \{x_0,\dots x_n\}$ be a partition of $[a,b]$ of mesh size less than $\delta.$ Define $\varphi$ on $[a,b]$ to be the piecewise linear function connecting the points $(x_0,f(x_0)), \dots, (x_n,f(x_n)).$ Use the MVT to see that on the interior of the $k$th subinterval we have

$$\varphi ' = \frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} = f'(c_k)$$

for some $c_k \in (x_{k-1},x_k).$ Thus on the interior of the $k$th subinterval, $|f'-\varphi '| < \epsilon$ by the way we chose the mesh size of $P.$ Thus we can say $|f'-\varphi '| < \epsilon$ on $[a,b]\setminus P.$

Now to the problem on $\mathbb {R}:$ Given $\epsilon >0,$ we can do the above on each $[n,n+1], n \in \mathbb {Z},$ obtaining a $\varphi_n$ on each $[n,n+1].$ Paste the $\varphi_n$ together to obtain $\varphi$ on $\mathbb {R}$ with $|f' -\varphi '|<\epsilon$ on $\mathbb {R}$ minus all of the partition points. This $\varphi $ has the desired properties for this $\epsilon,$ and we get the desired sequence by choosing a sequence of $\epsilon$'s $\to 0^+.$

Answer 3

It was already answered to your question (and brilliantly). I use this space only for some variation on the theme.

Your post reminds me of a method of proving the existence of an antiderivate of a continuous function.

Let $f$ be continuous on $[a,b]$. Now it is well known there exists a sequence of piecewise constant functions that converges uniformly to $f$.

One can show that, "antidifferentiating termwise", it is obtained a sequence of piecewise affine functions that converges to an antiderivative of $f$.

By "antidifferentiating termwise" I mean:

To find, for every approximating piecewise constant function, the generic antiderivative of every constant function defining it. To choose then the arbitrary constants so that $\,$i) the piecewise affine function so obtained is continuous on $[a,b]$ $\,$ii) all the piecewise affine functions of the sequence have the same value at a point of $[a,b]$. (one can do that of course)

Now if your $\phi'$ is my $f$ $\dots$

All this for the case $\phi \in C^1([a,b])$ which is the essential point of start.