📂Distribution Theory

Differentiability of Distributions is Continuous with Respect to Weak Convergence

Theorem¹

The differentiation of a distribution is continuous with respect to weak convergence. In other words, if $T_{k}$ weakly converges to $T$, then $\partial ^{\alpha} T_{k}$ weakly converges to $\partial ^{\alpha}T$.

$$ T_{k} \to T \quad \text{weakly} \implies \partial ^{\alpha} T_{k}\to \partial ^{\alpha}T \quad \text{weakly} $$

Here, $\alpha$ is any multi-index.

Explanation

It means that the differential operator of a distribution satisfies the condition for being continuous with respect to weak convergence.

$$ \lim \limits_{n\to \infty} f(p_{n})=f(p),\quad \forall \left\{ p_{n} \right\} \text{s.t.} \lim_{n \to \infty}p_{n}=p $$

The significance of this fact for distributions is that it does not hold for pointwise convergence or uniform convergence.

$$ f_{k} \to f \quad \text{pointwise or uniformly} \not \!\!\! \implies f^{\prime}_{k}\to f^{\prime} \quad \text{pointwise or uniformly} $$

That is, for a sequence of functions $f_{k}$ that converges to $f$, it is not guaranteed that $f^{\prime}_{k}$ converges to $f^{\prime}$. However, in the case of weak convergence, this is guaranteed.

Proof

Let’s assume that the sequence of distributions $T_{k}$ weakly converges to $T$. By definition of the differentiation of distributions, for any test function $\phi$, the following holds:

$$ \left( \partial ^{\alpha} T_{k} \right)(\phi) = \left( -1 \right)^{\left| \alpha \right| }T_{k}(\partial ^{\alpha} \phi) $$

Taking the limit $\lim \limits_{k \to \infty}$ on both sides, by assumption, we have:

$$ \begin{align*} \lim \limits_{k \to \infty}\left( \partial ^{\alpha} T_{k} \right)(\phi) &= \lim \limits_{k \to \infty}\left( -1 \right)^{\left| \alpha \right| }T_{k}(\partial ^{\alpha} \phi) \\ &= \left( -1 \right)^{\left| \alpha \right| }T(\partial ^{\alpha} \phi) \\ &= \left( \partial ^{\alpha}T \right)(\phi) \end{align*} $$

Therefore, the following holds:

$$ \partial ^{\alpha} T_{k}\to \partial ^{\alpha}T \quad \text{weakly} $$

Thus, if $T_{k}$ weakly converges to $T$, then $\partial ^{\alpha} T_{k}$ weakly converges to $\partial ^{\alpha}T$.

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