I am currently reading Hui Hsiung Kuo's book "Gaussian Measures in Banach Spaces" and there is an exercise (Exercise 21, p. 86) in which you are asked to show that for $1 \leq p < 2$, $(i, L^{2}[0,1], L^{p}[0,1])$ is not an abstract Wiener space (I have provided a definition of abstract Wiener space below). Here, $i$ denotes the inclusion map $i \colon L^{2}[0,1] \hookrightarrow L^{p}[0,1]$. One of the consequences of the definition of an abstract Wiener space is that the embedding map $i$ has to be compact, i.e. $L^{2}[0,1] \hookrightarrow \hookrightarrow L^{p}[0,1]$. As I am trying to show that it isn't an AWS, one of my first thoughts was to try to prove the embedding is not compact. But I could not produce a proof nor find one so far.
So my question is twofold:
-
Is the embedding $L^{2}[0,1] \hookrightarrow L^{p}[0,1]$ ($1 \leq p < 2$), or maybe more generally $L^{q}(S, \mathcal{S}, m) \hookrightarrow L^{p}(S, \mathcal{S}, m)$ ($1 \leq p < q$) a compact operator? Let us say that $(S, \mathcal{S}, m)$ is a finite measure space so the embedding makes sense.
-
If not, how do I prove that the triple $(i, L^{2}[0,1], L^{p}[0,1])$ is not an abstract Wiener space?
Best Regards,
Andre
Appendix (Definition of abstract Wiener space):
Kuo defines an abstract Wiener space starting from a (real) separable Hilbert space $(H, | \cdot |)$. Take another norm (if it exists) $\| \cdot \|$ on $H$ that is "measurable" in $H$, by which Kuo means (Definition 4.4, p. 59):
$\forall \varepsilon > 0 ~ \exists P_{\varepsilon} \in FP$ such that
$$
\mu \{ \| P x \| > \varepsilon \} < \varepsilon \quad \forall P \in FP, ~P \perp P_{\varepsilon},
$$
where $FP$ denotes the set of all finite-dimensional orthogonal projections on $H$. $\mu$ denotes the "standard" Gauss measure in $H$ (which is not $\sigma$-additive), defined on cylinder sets
$$
E_{P,F} = \{ x \in H ~|~Px \in F \}, \quad P \in FP, F \in B(P(H))
$$
(where the range $P(H)$ is finite-dimensional with dimension $\dim P(H) = n$ since $P$ is a finite-dimensional projection) via
$$
\mu(E_{P,F}) := \left( \frac{1}{\sqrt{2 \pi}} \right)^{n} \int_{F} e^{-\frac{|x|^2}{2}} dx.
$$
Then we define the Banach space $B$ as the completion of $H$ w.r.t. this norm $\| \cdot \|$. Obviously, $H$ is embedded into $B$ and we call the embedding map $i$. The triple $(i, H, B)$ is then called an abstract Wiener space. One of the consequences of this definition (Lemma 4.6, p.70 in the book) is that the embedding $i$ must be compact.
Best Answer
The embedding $L^2[0,1]\hookrightarrow L^p[0,1]$ is not compact. While it can be a little tricky checking that a given sequence has no convergent subsequence, there is another way: if $X\hookrightarrow Y$ and $x_n\rightharpoonup x$ weakly in $X$, then $x_n\to x$ in $Y$. (Can you prove this?) Now $L^2[0,1]$ is a Hilbert space, so $e_n\rightharpoonup 0$ weakly for any orthonormal sequence $\{e_n\}$ by Parseval's identity. If we take
$$h_n(x):=\begin{cases} 1&\text{if }\lfloor2^nx\rfloor\text{ is even,}\\ -1&\text{if }\lfloor2^nx\rfloor\text{ is odd} \end{cases}$$
then $\{h_n\}$ is an orthonormal sequence (to show this, note that if $m>n$, then for any interval $I$ on which $h_n$ is constant, $h_m$ splits $I$ into $2^{m-n}$ equally sized intervals with $h_m=1$ on half and $h_m=-1$ on the other, so $\int_Ih_nh_m\,\mathrm dx=0$). If $L^2[0,1]\hookrightarrow L^p[0,1]$ were compact, then we would have $h_n\to0$ in $L^p$. But $\|h_n\|_p=1$ for all $n$, so the embedding is not compact.