# Inner product in Sobolev spaces

Many books state the following:

Let $u, v \in W^{1,2}(\Omega)$ (a Sobolev space), the scalar product is:

$$u \cdot v = \int_{\Omega} uv \; dx + \int_{\Omega} \nabla u \nabla v \; dx$$

with $x = (x_1,x_2)$

Is that a definition or it is derived from any relationship? If is a relationship how to find it?

Is that also an inner product?

• Yes, it's an inner product. Dec 3, 2017 at 14:53
• This ought to be part of the definition of $W^{1,2}(\Omega)$ as a Hilbert space...a more proper definition would be something like "the Sobolev space $W^{1,2}(\Omega)$ admits the structure of a Hilbert space with inner product..."
– user113529
Dec 3, 2017 at 15:07
• But the Hilbert space does not requires the first derivative, that inner product "definition" is particular to Sobolev spaces, right? .
– Lin
Dec 3, 2017 at 15:12
• For sure it is peculiar to the Sobolev $W^{1,2}$. As any inner product $\langle, \rangle:H\times H \to \mathbb{K}$ is peculiar to $H$. Dec 3, 2017 at 16:42
• One could also imagine that $W^{1,2}$ is defined as a normed space with the norm $\| u \|^2_{W^{1,2}} := \| u \|^2_{L^2} + \| \nabla u \|^2_{L^2}$. Then the statement this norm is induced by the following scalar product (...) is a non-trivial (although very easy) observation. To check it, one just needs to check bi-linearity of $u \cdot v$ and its compatibility with the norm (i.e. $u \cdot u = \| u \|^2_{W^{1,2}}$). Dec 11, 2017 at 23:24

One way (there are others) to analyze solutions in modern PDE analysis is to:

1. Specify a space of functions with certain regularity (e.g. all functions on a bounded domain that have continuous derivatives up to second order)
2. Imagine a PDE as an operator on this function space.

We then use tools from functional analysis to say something about whether a solution exists (maybe play some game about examining how sequences in the domain get mapped to the range under the operator defined in 2).

We need to be careful about the spaces and operators we define if we want to play this game though. Consider our first guess for how we might want to play:

1. Consider continuous functions on a bounded interval under the sup norm
2. Consider the derivative operator

If we consider the sequence $f_n(x)=\sin(nx)$, The derivative operator will give the following bound: $$f'_n(x) = n \cos(nx) \\ ||f'_n|| \leq n ||f_n||$$ Which is to say, the derivative operator is not a continuous operator (I skipped a step, but stated the punchline). Continuous operators over spaces are things we need in order to play our game though -- think about what they do: if we have a sequence in the domain, a continuous operator allows us to say something about how the limiting object in the domain passes to the range.

So we'd like to manufacture complete spaces of functions over which we can define continuous maps. Sobolev spaces $W^{k,p}$ are one technology we have for this. What we do: control (via $L^p$ norm) how badly the first $k$ derivatives of a function can behave. These norms are created so that we can create continuous operators over these spaces.

That $W^{1,2}$ is Hilbert is coincidental, in general these things are Banach spaces.

• "Coincidental" is not really accurate. The fact that some Sobolev spaces are Hilbert spaces is hugely important; in many books and papers one finds the abbreviated notation $H^s$ for $W^{s,2}$ because for their purposes, only function spaces with Hilbert space structure are needed; $W^{s,p}$ with $p\ne 2$ never appear.
– user357151
Dec 3, 2017 at 19:15