Differentiable approximation $f(x) = x$ for $x>0$ and $0$ otherwise.

Question

I would like to find a twice continuously differentiable approximation of

$$f(x)= \begin{cases} 0 & x\leq 0 \\ x & x>0. \\ \end{cases}$$

Are there any approximations which are not defined piece-wise? Ideally, I'd like an approximation that will allow me to control the error given some norm.

Answer 1

Try $$ f(x) = \frac{\ln(1+e^{Kx})}{K} $$ for some largeish value of $K$.

Answer 2

Consider $$f(t)= \left\{ \begin{array}{cl} 0& \mbox{si $t\le0$}\\ \exp(-1/t)& \mbox{si $t>0$} \end{array} \right. $$ and for any $\delta>0$ set $$ g_\delta(t)=\frac{\textstyle f(t)}{\textstyle f(t)+f(\delta-t)}. $$ Then choose $\varepsilon>0$ and define $h(t)=tg_\delta(t-\varepsilon)$. One gets:

But I don't know whether this is what you didn't want... due to the piecewise definition of the first $f$.

Answer 3

For a general method of approximation, the convolution is really usefull. Take a positive function of $C^{\infty}_c([-1,1])$, as an exemple,

$$g(x) = \exp\left({-\frac{2}{(x-1)(x+1)}}\right)$$

Let's normalise it :

$$g_{norm}(x) = \frac{g(x)}{\int_{-1}^1 g(t) dt}$$

Then we define

$$g_{\epsilon}(x) = \frac{1}{\epsilon} g_{norm}(\frac{x}{\epsilon})$$

Then, for every $f \in L^1_{loc}$ you have that $g_{\epsilon} \star f \to f$ uniformly and that $g_{\epsilon} \star f \in C^{\infty}$

You also have $g_{\epsilon} \star f \to f$ in all $L^p$.

But I don't remember the error estimate...