problems with a delta function identity

Question

I want to prove $x\delta'(x)=-\delta(x)$. What I did was integrating the right side around 0 (since both sides are equals when $x \neq 0$ trivially):

$$\int_{-\varepsilon}^{\varepsilon}x\delta'(x)dx=\left[x\delta(x)\right]_{-\varepsilon}^{\varepsilon}-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx=0-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx=-\int_{-\varepsilon}^{\varepsilon}\delta(x)dx$$

I see the right and left sides are equal under this integration, but how does that tell me anything about their value at 0? Couldn't the integral be equal around 0 in "as small interval as we wish" yet still the original functions get different values at zero?

Answer 1

Answering your question requires that we answer the following: if we have a (generalized) function $d(x)$, then how do we know that $d(x)$ is the delta-function? In other words, what is the definition of the delta function?

One definition is that corresponding to the sifting property: $\delta(x)$ is the unique function $d(x)$ for which $\int_{-\infty}^\infty f(x) d(x) = f(0)$ holds for every function $f$. With this definition, if we want to show that $d(x) = -x\delta'(x)$ is the delta function, then it suffices to note that $$ \begin{align} \int_{-\infty}^\infty f(x)d(x)\,dx &= \int_{-\infty}^\infty f(x)(-x\delta'(x))\,dx \\ & = -\int_{-\infty}^\infty (xf(x))\delta'(x)\,dx \\ & = \int_{-\infty}^\infty (xf(x))'\delta(x)\,dx \\ & = \int_{-\infty}^\infty (xf'(x) + f(x))\delta(x)\,dx = 0\cdot f'(0) + f(0) = f(0). \end{align} $$ It follows that $d(x) = -x \delta'(x)$ is indeed equal to the delta function.

Note that in the above computation, I use the defining feature of the "function" $\delta'$: for any function $f(x)$, $\int_{-\infty}^\infty f(x)\delta'(x)dx = -f'(0)$.

Answer 2

As in @BenGrossman's comments and answer, your computation does capture the essence of the issue, though your concern about pointwise values is heading off into trouble.

For that matter, although the manipulation of integrals (as in your heuristic or in BenGrossman's explanation) tells the story, it is also possible to write an analogous, but/and absolutely rigorous argument, with fewer symbols: for a test function $f$, by unabashedly thinking of distributions as (continuous, linear) functionals on test functions: $$ (-x\delta')(f) \;=\; -\delta'(xf) \;=\; \delta((xf)') \;=\; \delta(f + xf') \;=\; f(0) + (xf')(0) \;=\; f(0)+0 \;=\; \delta(f) $$