Find the derivative of a piecewise constant function

Question

I want to differentiate the given function:

\begin{equation} f(x)= \begin{cases}1 \ \ \ \ \ x\in[0,1] \\ 0 \ \ \ \ \ \ x\not\in[0,1] \end{cases} \end{equation}

However, the simple $f'(x)=0$ seems a little bit too simple, since this is similar to the Dirac delta function, just stretched over the interval $[0,1]$ instead of at $x=0$.

So I tried to represent the function as a generalized function,

$$f(x)=\int_0^1\delta(x)\phi(x)\text{d}x,$$ where $\phi(x)$ is a locally summable function, infinitely differentiable, $\in C^\infty$.

and use the differentiation rule $$(D^\alpha f,\phi)=(-1)^{|\alpha|}(f,D^{\alpha}\phi),$$ with $\alpha=1$:

$$(D^1 f,\phi)=(-1)^{|1|}(\delta(x),D^{1}\phi)$$

$$(D^1 f,\phi)=-1(\delta(x),\phi'(x))$$

However, I am not sure how to continue. Any hints?

Thanks

Answer 1

Notice that just from the definition of the derivative, $f'(x) = 0$ for all $x\neq 0, 1$. In particular, for any $φ\in C^\infty_c(\mathbb{R})$, $$0 = \int φ(x)f'(x) dx = -\int φ'(x)f(x) dx = -\int_0^1 φ'(x) dx.$$ Choosing any test function $φ$ such that the last integral is nonzero (there are plenty such functions) gives a contradiction. Hence $f$ is not (weakly) differentiable.

Answer 2

The exact answer to this question is actually as follows. I was corrected by my Professor, and this is apparently the right procedure.

\begin{equation} f(x)= \begin{cases}1 \ \ \ \ \ x\in[0,1] \\ 0 \ \ \ \ \ \ x\not\in[0,1] \end{cases} \end{equation}

We use the formula

$$(f',\phi) = -(f,\phi') = -\int_\mathbb{R} f\phi' dx = -\int_{-\infty}^{x_1}f\phi' dx - \int_{x_1}^{x_2}f\phi' dx - \cdots -\int_{x_m}^{\infty}f\phi' dx $$

where $x_1=0$ and $x_2=1$, hence we obtain

$$(f',\phi) = -(f,\phi') = -\int_\mathbb{R} f\phi' dx = -\int_{-\infty}^{0}f\phi' dx - \int_{0}^{1}f\phi' dx - \int_{1}^{\infty}f\phi' dx $$

Now using integration by parts for each term, we get:

$$(f',\phi)=\int_{-\infty}^{\infty} \frac{df}{dx}\phi(x) dx + \sum_{j=1}^m [f(x_{j^+})-f(x{_j^-})]\phi(x_j)=$$

We have that $m=2$ so we obtain $$=\int_{-\infty}^{\infty}\frac{df}{dx}\phi(x) dx + [f(x{_1^+})-f(x{_1^-})]\phi(x_1)+ [f(x{_2^+})-f(x{_2^-})]\phi(x_2)$$ which is $$= [f(0^+)-f(0^-)]\phi(0)+ [f(1^+)-f(1^-)]\phi(1)$$

which gives, since $\delta(x)$ is normally centered at $x=0$ hence, with $\phi(1)=\delta(x-1)$, we obtain $$(f',\phi)=\delta(x)+ \delta(x-1)$$

Note that this is only possible since $\phi(x) \in \mathscr{L}_{loc}^\infty \to \delta(x)$ as $k\to \infty \in \mathbb{R}$, which is satisfied in the given space of the function.