The product of two distributions is not a distribution Suppose $u(x)$ is the Heaviside function, which takes value $1$ when $x\ge0$ and takes value $0$ when $x<0$. Then the derivative of $u$ is a delta function $\delta_0$, or a distribution.
Now my question is can we properly define the integral:
$$\int_{-1}^1x^2|u'(x)|^2dx.$$ 
If there is no square inside the integral, i.e., $\int_{-1}^1xu'(x)dx$ is easy to decide.
But for the square case, I'm more inclined to the negative answer, and typically in some Functional Analysis textbooks, it is always remarked that the product of two distributions are generally not a distribution, but none of them provide a proof. 
So here I post this question, hope those have a profound understanding of functional analysis could help me.
Any advice is appreciated.
 A: Distributions are correctly thought of as continuous maps $F:\mathcal D\to\mathbb C$ where $\mathcal D$ is "test functions," or smooth functions with compact support in some region $\Omega$. In light of this abstraction, they do not enjoy all of the same operations that you have on traditional functions, although many extend by self-adjointness. For example, consider multiplication by a smooth function $\psi$. Then for $f\in L_{loc}^1$ and $\varphi\in\mathcal D$,
$$\int (\psi f)\varphi=\int f(\psi\varphi).$$
Therefore, extrapolating from the case where $F(\varphi)=\int f\varphi$, we can DEFINE $(\psi F)(\varphi)=F(\psi\varphi)$ since $\psi\varphi\in\mathcal D$. The general idea is we identify some duality in the special case where $F$ is given by integration against a function, then we use that duality to define the corresponding operation on distributions. The same thing can be done to define for distributions derivatives, Fourier transforms, etc.
Suppose we tried to do something similar to define the product of distributions. In the case where $F,G\in\mathcal D'$ are given by $f,g\in L_{Loc}^1$, we would want, for any $\varphi\in\mathcal D$,
$$(FG)(\varphi)=\int fg\varphi=\int f(g\varphi)=F(g\varphi)$$
but $g\varphi$ is probably not smooth since $g$ is not smooth, so you cannot plug it into $F$. On top of that $G$ might not even by given by a function $g$! This is just an example of a difficulty that arises when you try to define what it would be to multiply distributions. I suppose the reason you haven't seen a formal proof is because the difficulty is coming up with a definition that agrees with our idea of multiplication--so what would you actually prove?
Hopefully this begins to clear up your question about the Heaviside function as well. As you know, in the sense of distributions, $u'=\delta$ so $(u')^2=\delta^2$ is not defined. One could go ahead and define some meaning for powers of Dirac functions; this is fine but it would not be consistent with the general theory of distributions.
A: Rather than asserting that the "product of two distributions is generally not a distribution" the textbook I'm learning from (Hunter & Nachtergaele - Applied Analysis (2001)) says it is not possible to define a product with the same algebraic properties as the pointwise product of functions (pg 295). They do not prove this, but rather state the problem precisely as an exercise (11.7) and provide a hint which I found very enlightening: compute the product $x \cdot \delta(x) \cdot \text{p.v.}(1/x)$
where $x$ is interpreted as a distribution, and $\text{p.v.}(1/x)$ is the distribution defined by
$$ \text{p.v.}(1/x)(\varphi) = \lim_{\epsilon \to 0} \int_{|x|>\epsilon} \frac{1}{x} \varphi(x) \ dx.$$
The counter example amounts to
\begin{align*}
\delta \cdot (x \cdot \text{p.v.}(1/x)) &= 1 \cdot \delta = 1 \\
(\delta \cdot x) \cdot \text{p.v.}(1/x) &= 0 \cdot \text{p.v.}(1/x) = 0
\end{align*}
where every multiplication is a multiplication of a distribution by a smooth function of polynomial growth which is reasonable: if $f$ is smooth and of polynomial growth, and $T$ is a distribution, define the product as the distribution
$$
(fT)(\phi) = T(f \cdot \phi)
$$
where $\cdot$ is pointwise multiplication and is well defined for smooth functions.
In effect, we cannot create a product operator that is consistent with pointwise multiplication for smooth functions, and has the usual properties of a multiplication operator (specifically, associativity in this counter example).
A: Suppose $\phi(x)$ is a smooth function on $R$ and has compact support on $[-1,1]$. In addition, we require this $\phi(x)$ satisfy:
$$\phi(x)\ge0,\qquad\int_{-1}^1\phi(x)dx=1,\qquad\int_{-1}^1|\phi(x)|^2dx\neq 0.$$
One could easily check that such choices of $\phi(x)$ exist.
Next let $\delta$ denote the delta function of $0$, or the $0$-distribution
, or equivalently the weak derivative of Heaviside function. Then we define $\delta_n(x)=n\phi(nx)$. Immediately one could check that 
$$(\delta*\delta_n)(x)=\delta_n(x)=n\phi(nx).$$
Now for any $\psi(x)\in \mathcal{D}(R)$,
$$<(\delta*\delta_n)^2,\psi)>=\int_{-1}^1n^2|\phi(nx)|^2\psi(x)dx.$$
Suppose $\psi(x)$ is identical $1$ in some $0$-neighborhood, then
$$<(\delta*\delta_n)^2,\psi>=n\int_R|\phi(x)|^2dx\to\infty\quad as\quad n\to\infty .$$
Which means, in the above sense $(\delta*\delta_n)^2\to\delta^2$ as $n\to\infty$ is not a properly defined distribution.
