Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a follow-up from this question. It was proved that, if $X$ is a linear subspace of $\mathbb{L}^1 (\mathbb{R})$ such that:

  • $X$ is closed in $\mathbb{L}^1 (\mathbb{R})$;
  • $X \subset \bigcup_{p > 1} \mathbb{L}^p (\mathbb{R})$,

then $X \subset \mathbb{L}^p (\mathbb{R})$ for some $p>1$.

I was wondering whether one could find a subspace $X$ satisfying these hypotheses and which is infinite-dimensional. It turns out this is possible. If one chooses a bump function, and considers the closure for the $\mathbb{L}^1 (\mathbb{R})$ norm of the space generated by the translates by integers of this bump function, one can emulate the $\ell^1$ space. The resulting $X$ will be closed, and included in $\mathbb{L}^p (\mathbb{R})$ for all $p>0$. To avoid this phenomenon, I'll restrict myself to smaller spaces.

Is there a linear, closed, infinite-dimensional subspace $X$ of $\mathbb{L}^1 ([0,1])$ which is included in $\mathbb{L}^p ([0,1])$ for some $p>1$?

The problem is that any obvious choice of countable basis will very easily generate all of $\mathbb{L}^1 ([0,1])$ (polynomials, trigonometric polynomials...), or $\mathbb{L}^1 (A)$ for some $A \subset [0,1]$, or at least one function which is in $\mathbb{L}^1$ but not in $\mathbb{L}^p$ for $p>1$...

share|cite|improve this question
up vote 3 down vote accepted

An infinite-dimensional example can be obtained as follows:

Let $(Y_n)_{n=1}^\infty \subset L^2[0,1]$ be a sequence of independent standard Gaussians, that is random variables with density $\frac{1}{\sqrt{2\pi}}e^{-t^2/2}$.

Let $X$ be the closed linear span of $(Y_n)_{n=1}^\infty$ in $L^2[0,1]$. I claim that $X \subset L^p$ for all $1 \leq p \lt \infty$.

Consider a finite linear combination $S_N = \sum_{n=1}^N a_n Y_n$. Then $S_N$ is a normal random variable, has mean zero and variance $\sigma^2 = E(|S_N|^2) = \sum_{n=1}^N |a_n|^2$, so $\frac{1}{\sigma} S_N$ is a standard Gaussian, too.

This shows in particular that the space $X$ is isometrically isomorphic to $\ell^2$.

Moreover, we can compute for the $L^p$-norm of $S_N$ as above that $$ \begin{align*} E(|S_N|^p) & = \sigma^p \frac{2}{\sqrt{2\pi}} \int_{0}^\infty t^p e^{-t^2/2}\,dt = \sigma^p \frac{2}{\sqrt{2\pi}} 2^{(p-1)/2} \int_{0}^\infty s^{(p-1)/2} e^{-s}\,ds \\ &= \sigma^p \sqrt{\frac{2^p}{\pi}} \Gamma\left(\frac{p+1}{2}\right), \end{align*} $$ so $\|S_N\|_p = C_p \cdot \|S_N\|_2$ for all $1 \leq p \lt \infty$.

This shows that $X$ is a closed subspace of all spaces $L^p[0,1]$, and up to a constant factor depending only on $p$, its norm is the same as the $L^2$-norm.

share|cite|improve this answer
A reference containing expositions of the techniques used in both answers is Albiac-Kalton, Chapter 6. See in particular Proposition 6.4.13 for the approach via Gaussians and section 6.2 for the Rademacher techniques and the Khintchine inequality and (some of) its uses. See also this answer for an application of these ideas. – t.b. Jul 14 '12 at 16:10
These are two very nice answers; it's hard to choose between them, but I'm a bit partial to arguments with a probability flavor. Also, thank you for the references. – D. Thomine Jul 14 '12 at 16:54

There exists such space. Let $\{R_i\}$ be the set of Rademacher functions on $[0,1]$. Specifically, let $$R_k(x) = sign(\sin(\pi k x)).$$ For every $\alpha = (\alpha_1, \dots, \alpha_k, \dots) \in \ell_2$, let $f_{\alpha}(x) = \sum \alpha_k R_k(x)$. Note that by the Khintchine inequality, for every $p\in [1,\infty)$, $$A_p \|\alpha\|_2 \leq \|f_{\alpha}\|_p \leq B_p \|\alpha\|_2$$ (where $A_p$ and $B_p$ depend only on $p$). In particular, $\|f_{\alpha}\|_1 < \infty$ for all $\alpha\in \ell_2$. Therefore, $f_{\alpha} \in L_1[0,1]$ for every $\alpha \in \ell_2$. Consider $V = \{f_{\alpha}: \alpha\in \ell_2\}$.

Clearly, $V$ is a linear subspace of $L_1[0,1]$. Moreover, $V$ is a closed subspace. Indeed if a sequence $f_{\alpha^{(1)}},f_{\alpha^{(2)}}, f_{\alpha^{(3)}}, \dots$ converges to $g\in L_1[0,1]$, then $f_{\alpha^{(k)}}$ is a Cauchy sequence in $L_1[0,1]$. Therefore, by the Khinchine Inequality, $\alpha^{(k)}$ is a Cauchy sequence in $\ell_2$. Hence it converges to some $\alpha^* \in \ell_2$. Then $f_{\alpha^{(k)}} \to f_{\alpha^*}$, and $g = f_{\alpha^*}$ (a.e.).

The Khintchine inequality implies that $V$ is a subspace of every $L_p[0,1]$ (where $1\leq p < \infty$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.