1

It is well known that every closed subspace of a Banach space is itself a Banach space. However this only addresses one direction, and the subspace norm is the one it inherits from the larger space. But what if $X_1\subset X$ is a Banach space under norm $\|\cdot\|_1$, where $X$ is a Banach space under a different norm $\|\cdot\|$? Can we say anything about whether $X_1$ is closed or dense in $X$?

An example would be $L^{p_1}(I)\subset L^{p_2}(I)$, where $I$ is a measure space, and $0<p_1<p_2<\infty$. Is $L^{p_1}(I)$ closed or dense in $L^{p_2}(I)$, under the norm $\|\cdot\|_{p_2}$? Is the $\|\cdot\|_{p_1}$ topology even the same as the $\|\cdot\|_{p_2}$ one, on subspace $L^{p_1}(I)$?

Jean Marie
  • 81,803
user760
  • 1,169
  • If $(X,|;|_X)$ and $(Y,|;|_Y)$ are Banach spaces, $X\subset Y$, the inclusion $\iota:X\rightarrow Y$, $x\mapsto x$, is bounded, and $X$ is closed in $(Y,|;|_Y)$, then the norms $|;|_X$ and $|;|_Y$ are equivalent and yield the same topology on $X$. This follows as a consequence of the open map theorem. – Mittens Nov 04 '22 at 17:04
  • 1
    @OliverDíaz Actually I do not feel the nature of this problem. There must be some relation between the norms restricted to $X. $ Otherwise anything may occur as has been pointed out in the answer below. – Ryszard Szwarc Nov 04 '22 at 19:55
  • 1
    @RyszardSzwarc: My goal was/is to give examples where (using the notation in my first comment) $X$ is not either closed nor dense in $(Y,|;|)Y)$ (even if the inclusion is bounded). Of course, if there are not relations between no norms (the inclusion is not bounded for example) all gloves are off (anything can happen). – Mittens Nov 04 '22 at 20:57

1 Answers1

1

  • In genereal, if $(X,\|\;\|_X)$ and $(Y,\|\;\|_Y)$ are Banach spaces, $X\subset Y$, $X$ is closed as a subset of $(Y,\|\;\|_Y)$ and the inclusion map $\iota:X\rightarrow Y$ given by $x\mapsto x$ is bounded, then the norms $\|\;\|_X$ and $\|\;\|_Y$ on $X$ are equivalent and thus, yield the same topology. Further, there are constant $a,b>0$ such that $$a\|x\|_X\leq \|x\|_Y\leq b\|x\|_Y,\qquad x\in X$$

  • The spaces $(\ell_1,\|\;\|_1)$ and $(\ell_\infty,\|\;\|_\infty)$ are Banach spaces. Clearly $\ell_1\subset\ell_\infty$, however $\ell_1$ is neither dense nor closed in $(\ell_\infty,\|\;\|_\infty)$. Indeed, the constant function $\mathbb{1}$ is not in the closure of $\ell_1$ under $\|\;\|_\infty$; each element of the sequence of functions $\mathbf{x}_n$ given by $\mathbf{x}_n(m)=\frac{1}{\sqrt{m}}\mathbb{1}_{\{1,\ldots,n\}}(m)$ is in $\ell_1$ and converges in $\ell_\infty$ to the function $\mathbf{x}(m)=\frac{1}{\sqrt{m}}$ which is not in $\ell_1$.

  • As pointed out by others, including the OP, if $(X,\mathscr{B},\mu)$ is a measure, $L_s(\mu)\subset L_r(\mu)$ for some (and hence all) $0<r < s\leq\infty$ iff $\mu(X)<\infty$. Then, if $(X,\mathscr{B}, \mu))$ is a finite measure space, the inclusion $\iota: L_s(\mu)\rightarrow L_r(\mu)$ is bounded by Holder's inequality. As $L_\infty(\mu)\cap L_s(\mu)$ contains space of simple functions, $L_\infty(\mu)\cap L_s(\mu)$ is dense in $L_r(\mu)$. If $X$ is not finite then the inclusion $L_s(\mu)\subset L_r(\mu)$ for $r<s$ is proper.

  • In a more general setting ($X$, $Y$ are infinite dimensional Banach spaces), things are much more complicated. See this posting and other links associated to it.

  • Here is one interesting example in $L_p(\mu)$ spaces. Suppose $(X,\mathscr{F},\mu)$ is a probability example that admits an i.i.d sequence $(\xi_n:n\in\mathbb{N})$ of Bernoulli random variables taking values $\pm1$ with probability $1/2$. Let $$R=\{\sum_na_n\xi_n: \mathbf{a}=(a_n)\in\ell_2(\mathbb{N})\}$$ This is a subspace of $L_2(\mu)$ which is isometric isomorphic to $\ell_2(\mathbb{N})$ and so $R$ is closed in $L_2(\mathbb{P})$. A theorem by Khintchine states that for any $0<p<\infty$ there are constants $k_p,K_p>0$ such that for any $\xi\in R$ $$k_p\|\xi\|_p\leq \|\xi\|_2\leq K_p\|\xi\|_p$$ This is an example of a thin Banach space in $L_2(\mu)$ that is also a closed subspace with $L_p(\mu)$ for all $0<p<\infty$. The condition $0<p<\infty$ cannot be extend to $p=\infty$ because $R$ is infinite dimensional.

Mittens
  • 39,145
  • I just deleted my answer because I realised a fatal flaw (my norm $|f|=|f(1)|$ was only a pseudonorm). Your first point strikes me as correct only if $X\subset Y$ is to be read as: “$X$ is a subspace of $Y$”. If $X\subset Y$ is true on the set level but the norms are arbitrary, there is no reason for the inclusion to be continuous. – FShrike Nov 04 '22 at 21:43
  • @FShrike: Yes, $X\subset Y$ is meant to say two things: $X,Y$ are both linear spaces with respect to the same field $(\mathbb{R}$ or $\mathbb{C}$), and as sets, $X\subset Y$ (this will imply that $X$ is a linear subset of $Y$. – Mittens Nov 04 '22 at 21:54
  • @FShrike: Notice that in the first bullet in my posting, the continuity if the inclusion is stated as assumption, not a conclusion. – Mittens Nov 04 '22 at 23:36
  • Can you elaborate on the sentence "If $X$ is not finite then the inclusion $L_s(\mu)\subset L_r(\mu)$ for $r<s$ is proper", in your 3rd bullet point?

    According to the Wikipedia page for $L^p$ spaces, for $1\leq p<q \leq \infty$, both directions of inclusion $L^p\subset L^q$ and $L^q\subset L^p$ are possible, depending on the underlying measure. This seems very surprising to me. I've always thought there's only 1 direction.

     – user760 Nov 06 '22 at 20:26