A continuous embedding of two Hilbert spaces X ↪ Y is of Hilbert Schmidt type if for every orthonormal basis φ j, j ∈ N, of X it holds that ∑ j = 1 ∞ ∥ φ j ∥ Y 2 < ∞.
For instance, the mapping x_0\mapsto \delta_{x_0} is a continuous embedding of X into the space of finite Radon measures on X, equipped with its vague topology.
Theorem 3.8 The mapping f → [ f ⋆ s n s n ] is a continuous embedding of S + ( R ) into H ( Y ).
end{aligned} Here ↪ stands for a continuous embedding.
Let Ω be a bounded domain with a Lipschitz boundary; there is a continuous embedding V ↪ L y for y ∈ [ 2, 2 n n − 2 ] when n ≥ 3, and y ∈ [ 2, + ∞ ) when n = 1, 2. Then there exists γ y > 0, such that ∥ u ∥ L y ≤ γ y ∥ u ∥, ∀ u ∈ V. (4).
If (0<|Omega|a continuous embedding (L^{p(x)}(Omega)hookrightarrow L^{q(x)}(Omega)).
(ii) If V satisfies the assumption (V), then there is a compact embedding (Xhookrightarrow L^{s}({mathbb {R}}^{N})) for any (s in[p, p^)). . There is a continuous embedding (W^{1,p}({mathbb {R}}^{N})hookrightarrow L^{s}({mathbb {R}}^{N})) for any (s in [p, p^]).
Notice that with a continuous embedding and moreover by Sobolev embedding theorem.
There is a continuous embedding (H_{T}^{1}hookrightarrow C([0,mathbb{R}^{}^{N})), and the embedding is compact.
If is Lipschitz continuous and, is measurable and satisfies a.e., then there is a continuous embedding.
