1
$\begingroup$

I have an increasing continuous function $f:{\mathbb R}\rightarrow {\mathbb R}$ which is not differentiable everywhere, and I would like to approximate it with an infinitely differentiable function $g\in C^{\infty}$. I found this paper:

Uniform approximation of continuous mappings by smooth mappings with no critical points on Hilbert manifolds. Arxiv,

which tells me I can do it uniformly. However, nothing tells me that $g$ is also increasing, and this property is essential in my context. I would appreciate any pointers on the existence of a uniform approximation to $f$ in $C^{\infty}$ which is also increasing.

P.S. My function is basically: $K_1 x I(x<0) + K_2x I(x\geq 0)$, $K_1,K_2>0$.

$\endgroup$
5
  • 2
    $\begingroup$ One can approximate the given continuous function by an increasing piece-wise affine function and then smooth it. $\endgroup$
    – Petya
    Commented Oct 30, 2014 at 13:46
  • $\begingroup$ Is your function actually piecewise linear with only two pieces? For that convolution against a compactly supported even mollifier will get you immediately your uniform approximation. So I assume there are some additional technicalities? $\endgroup$ Commented Oct 30, 2014 at 13:53
  • $\begingroup$ @WillieWong Uniformly approximating the linear function with two pieces with a smooth function is enough for me. $\endgroup$
    – Leffe
    Commented Oct 30, 2014 at 14:08
  • $\begingroup$ @Petya Thanks for your comment. For the linear function with two pieces, how can this smoothing be done? It appears that one could use the same linear functions. $\endgroup$
    – Leffe
    Commented Oct 30, 2014 at 14:14
  • $\begingroup$ Let $f$ be your two-piece linear function. Let $\varphi\in C^\infty_0((-\epsilon,\epsilon))$ for some small $\epsilon$, such that $\varphi$ is even, with integral $\int \varphi = 1$, and $x\varphi' \leq 0$. Then you can check that the convolution $\varphi*f$ is increasing, smooth, and agrees with $f$ outside $(-2\epsilon,2\epsilon)$. $\endgroup$ Commented Oct 30, 2014 at 14:17

1 Answer 1

1
$\begingroup$

Let $f$ be your two-piece linear function. Let $\varphi\in C^\infty_0((-\epsilon,\epsilon))$ for some small $\epsilon$, such that

  • $\varphi$ is even
  • the integral $\int \varphi = 1$
  • $x\varphi' \leq 0$

Then you can check that the convolution $\varphi*f$ is increasing, smooth, and agrees with $f$ outside $(-2\epsilon,2\epsilon)$.

Taking appropriately rescaled versions of $\varphi$ you get uniform approximations.


Sketch of Proof:

  1. Observe that since $\varphi$ is even $\int \varphi(x) x \mathrm{d}x = 0$, this guarantees that if $f$ is linear in $(x-\epsilon,x+\epsilon)$, $\varphi* f(x) = f(x)$ (write $f(y) = (f(y) - f(x)) + f(x)$.)
  2. Let $g(x) = \varphi*f(x)$, we have $g' = \varphi'*f$. In particular $g'(x) = \int_0^\epsilon \varphi'(y) ( f(x-y) - f(x+y)) \mathrm{d}y$. So using that $x\varphi'(x) \leq 0$ and $f$ is increasing you get that $g'(x) \geq 0$.
  3. It is a standard fact that $\varphi*f$ is smooth.
  4. Let $\varphi_\delta(x) = \frac1\delta \varphi(x / \delta)$. Observe that $$ \varphi_\delta* f(x) - f(x) = \int \varphi_\delta(y) (f(x - y) - f(x)) ~\mathrm{d}y $$ and observe that the difference is always zero when $|x| > 2\delta\epsilon$. You have that $$ |\varphi_\delta*f(x) - f(x)| \leq \sup_{x,y\in (-2\delta\epsilon,2\delta\epsilon)} |f(x) - f(y)| $$ which can be easily controlled by uniform continuity (of continuous functions on compact sets).
$\endgroup$
3
  • $\begingroup$ Also, can the difference between $f$ and $\varphi * f$ be as small as you want in $(-\epsilon,\epsilon)$ by choosing the right $\varphi$? $\endgroup$
    – Leffe
    Commented Oct 30, 2014 at 14:26
  • $\begingroup$ // References would be a bit hard. This is one of those standard things that you can mostly use without referencing. I've included a sketch of the proof above, and if you must really cite something, I guess you can cite this post. $\endgroup$ Commented Oct 30, 2014 at 14:35
  • $\begingroup$ For arbitrarily fine approximations, see point 4. of the sketch of proof above. $\endgroup$ Commented Oct 30, 2014 at 14:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .