Recently I saw an interesting lemma:

For any $s>0$, the closed unit ball in $H^s$ is also closed in the $L^2$ norm. That is, suppose $u_j\in H^s$ and $\|u_j\|_{H^s}\le 1$. Suppose $u_j\to u$ in $L^2$. Then $u\in H^s$ and $\|u\|_{H^s}\le 1$.

A possible proof of the above lemma is to take the Fourier transform and use Fatou's lemma. Another possible proof is to use Riesz representation theorem to find out successively higher derivatives of $u$ (seems to work only when $s$ is a positive integer, though).

Contemplating the above proofs, it seems to me that there is some sort of compatibility between $H^s$ and $L^2$ for the above lemma to hold. For a nonexample, it is easy to see that the closed unit ball in $C^0([0,1])$ is not closed in $L^2$ norm simply by taking a sequence of continuous functions converging in $L^2$ to $1_{[0,1/2]}$, for example.

That brings me to the question: Let $Y$ be a dense subspace of a Banach space $X$. What condition on the pair $(X,Y)$ makes the closed unit ball in $Y$ also closed in $X$? If there is no simple iff condition, is there any nontrivial sufficient/necessary condition? Among the common spaces in Analysis (e.g. Holder spaces or Sobolev spaces), is there a list of all the pairs for the above property to hold?

Thanks for your time.

  • $\begingroup$ I know the condition as a completeness lemma: If the unit ball of a normed space $X$ is closed in a Banach space $Y$ where $X$ is continuously embedded, then $X$ is also complete. A variant of this easy exercise to locally convex spaces is sometimes attributed to W. Robertson. $\endgroup$ Nov 23, 2015 at 13:17
  • $\begingroup$ One condition which follows directly from Hahn-Banach is that there be a family of elements of the dual of $X$ so that the ball of $Y$ is the intersection of the unit balls of the seminorms $|f|$. This suffices to show your result, also in the generalised form that the domain of definition of an (unbounded) self-adjoint operator on a Hilbert space, with its natural structure---also that of a Hilbert space---has your property. Applications: various Sobolev type space obtained by taking powers (not necessarily integral) of the Laplace or Schrödinger operator on Riemann manifolds. $\endgroup$
    – dalry
    Nov 24, 2015 at 15:22

2 Answers 2


Suppose that you have two Banach spaces $X$ and $Y$, and a (bounded) operator $TX:\to Y$. The operator $T$ is called semi-embedding if $T$ is injective and $T(B_X)$ is closed in $Y$. So your are asking when natural maps between classical Banach spaces are semi-embeddings.

I do not know characterizations of semiembeddings better than the definition but, for example, if $T:X\to Y$ has dense range, then the conjugate operator $T^*:Y^*\to X^*$ is a semi-embedding.

Semi-embeddings between Banach spaces have been studied because they preserve some isomorphic properties:

  • Bourgain and Rosenthal proved that if $X$ is separable, Y has the Radon-Nikodym property and there exists a semiembedding $T:X\to Y$, then $X$ has the Radon-Nikodym property.
  • Neidinger and Rosenthal proved that if $T:X\to Y$ is injective and every restriction of $T$ to a closed subspace of $X$ is a semiembedding, then $T$ is tauberian. In this case, many isomorphic properties pass from $Y$ to $X$. For example, if $Y$ is reflexive or contains no copies of $\ell_1$, then $X$ has the same property (See Neidinger's Ph. D. Thesis).

These results are useful to find examples of injective operators that are not semi-embeddings.

  • $\begingroup$ That at least answers my original question: Just take $X=L^2$ and $Y=H^{-s}$. Since the embedding $T: L^2\to H^{-s}$ has dense range, $T^*: (H^{-s})^*=H^s\to (L^2)^*=L^2$ is a semi-embedding. $\endgroup$
    – Fan Zheng
    Nov 22, 2015 at 17:54

Gonzalez pointed to some deeper results on semi-embeddings. When $X$ is reflexive, there is an elementary characterization: $B_Y$ is closed in $X$ iff there is a $Z$ s.t. $Y = Z^\ast$ isometrically and the inclusion map $J$ from $Y$ into $X$ is weak$^\ast$ to weak continuous. This is automatic if $Y$ is reflexive and cannot happen if $Y$ is not isometrically isomorphic to a dual space; in particular, if $Y$ is $C[0,1]$, $c_0$, or $L_1$.


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