What does : we take $f=(1-\Delta )^Nu$ in the distribution mean?

Question

What does : we take $f=(1-\Delta )^Nu$ in the distribution mean for a function $u$ ? I recall that $$(1-\Delta )^N=\left(1-\sum_{i=1}^n\partial _i^2\right)^N.$$ First what is $(1-\Delta )^N$ ? Is it $$\sum_{k=0}^\infty (-1)^k\Delta ^k \ \ ?$$ if yes, what would represent $\Delta ^k$ ? And finally, when we say "in the distribution sense", is it that the derivative is taking as the Sobolev definition ?

Answer 1

A distribution is linear functional on a space of nice functions, usually the space of compactly supported infinitely differentiable functions, $C_c^\infty$. The application is sometimes written $\langle u, \phi \rangle,$ where $u$ is the distribution and $\phi$ is the nice function.

The application of a linear operator $T$ on a distribution $u$ is normally defined by instead applying its adjoint $T^*$ on the nice function: $$\langle Tu, \phi \rangle = \langle u, T^*\phi \rangle$$ For $T = (1-\Delta)^N$ we have $T^* = T$.

However, the author probably doesn't think much about the actual distributional definition of the operator, but rather on how it practically works and that although the operator acts on a function, the result can be a distribution. For example, if a function $f$ has a discontinuous derivative, like $f(x) = |x|$, then its second derivative has a Dirac delta part, $f''(x) = 2 \delta(x)$ for the given example.