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I want to know that the meaning of the following. $$ W^{n,1}\textrm{ is continuously imbedded into }L^2$$ Here, $W^{n,1}$ is a Sobolev space.

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First Google hit (incidentally, it's "embedded", not "imbedded"). – David Mitra Apr 21 '12 at 20:25
up vote 3 down vote accepted

Recall that for some (open?) set $U \subset \mathbb{R}^n$, $L^2(U)$ is the set of functions where the integral

$$||f||_{L^2(U)} := \displaystyle\int_U |f|^2 dx < \infty.$$

The reason we introduce Sobolev spaces "$W^{n,k}$" is to account for partial differentiation in $n$ variables of functions already in $L^2$ (technically weak differentiation, but I digress) each $k$-times differentiable. So for $U \subset \mathbb{R}^n$, the Sobolev space $W^{n,1}(U)$ is the set of functions that are $1$ time differentiable in the $n$ variables of $\mathbb{R}^n$.

The norm $||f||_{W^{n,1}}$ is no longer just the integral squared, but rather takes into account the derivative term:

$$||f||_{W^{n,1}} = \left( ||f||_{L^2}^2 + ||Df||_{L^2}^2 \right)^{\frac{1}{2}}.$$

To say that "$W^{n,1}$ is continuously embedded in $L^2(U)$" means that $W^{n,1}$ is a subset of $L^2$ and that

$$||f||_{L^2} \leq C ||f||_{W^{1,n}}.$$

One of the very important reasons for it is that is continuous embedding of a normed vector space $X$ into a normed vector space $Y$ is one of the two conditions required to have $X$ compactly embedded into $Y$, which give you nice properties with respect to bounded sequences and convergent subsequences.

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This means that each $f\in W^{n,1}$ is an $L^2$ function (that is, there is a representative $\tilde{f}$ of $f$ such that $\tilde{f}\in L^2$) and that there is a constant $C>0$ (independent of $f$) such that $$ ||f||_{L^2} \le C ||f||_{W^{1,n}} $$ ($C$ will usually depend on $n$ and the domain in question and probably some other quantities.)

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