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., $\phi_(t/\lambda)/\lambda$.

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