Wednesday, April 04, 2007

Approximating not-nice functions by nice functions

In the last class I asserted that it was possible to approximate certain not-nice functions like $1_{t < t_0}$ and $|t|$ by $B$-nice functions. For example, I asserted that for all $t_0 \in \mathbb{R}$ and $0 < \lambda < 1/2$ there exists a function $\Delta_{t_0,\lambda} : \mathbb{R} \to \mathbb{R}$ which is $O(1/\lambda^4)$-nice and approximates the $t_0$-step-function in the following sense: $\Delta_{t_0, \lambda}(t) = 1$ for $t < t_0 - \lambda$; $\Delta_{t_0, \lambda}(t) = 0$ for $t > t_0 + \lambda$; and, $0 \leq \Delta_{t_0, \lambda}(t) \leq 1$ for $|t - t_0| \leq \lambda$.

There was a question in class as to whether this could really be possible. Specifically, if $\Delta_{t_0, \lambda}$ is $0$ for all $t > t_0 + \lambda$, and it's smooth, then it has all derivatives equal to $0$ for all $t > t_0 + \lambda$. Shouldn't this make it $0$ everywhere?

Well, not so. It's true that for any $t > t_0 + \lambda$, all derivatives are $0$. But Taylor's theorem does not force such a function to be $0$ everywhere. Remember that Taylor (for smooth functions) just says that for any $r \in \mathbb{N}$, $f(x + \epsilon) = f(x) + \epsilon f'(x) + \frac{\epsilon^2}{2!} f''(x) + \cdots + \frac{\epsilon^{r-1}}{(r-1)!} f^{(r-1)}(x) \epsilon^{r-1} + \frac{\epsilon^r}{r!} f^{(r)}(y)$ for some $y \in [x, x + \epsilon]$.

For a concrete example, consider the function $\phi(t)$ which is $\exp(-1/(1-t^2))$ for $|t| < 1$, and is $0$ for $|t| \geq 1$. It's easy to see that $\phi(t)$ is smooth on $(-1,1)$ and it's not hard to check that its derivatives, of all orders, at $\pm 1$ (when approached from inside) are $0$.

In fact, functions like $\phi$ (known as bump functions or mollifiers) are what one uses to create functions like $\Delta_{t_0, \lambda}$ -- essentially, by taking $\Delta * \phi_\lambda$, where $\Delta_{t_0}$ denotes the actual discontinuous step function, $*$ denotes convolution, and $\phi_\lambda$ denotes a compressed version of $\phi$, viz., $\phi_(t/\lambda)/\lambda$.

1 comment:

Unknown said...

I wish to convey my gratitude for your kindness for visitors who absolutely need assistance with that idea. Your real dedication to passing the message all over had been pretty practical and has without exception enabled associates much like me to realize their targets. Your own warm and friendly key points indicates a lot to me and extremely more to my mates. With thanks; from each one of us. Cara Meyembuhkan Kutil Kelamin dari Dokter Spesialis de Nature