Let $(W, \|\cdot\|_W)$ be a real separable Banach space equipped with a non-degenerate Gaussian Borel measure $\mu$. Let $H \subset W$ be the corresponding Cameron-Martin Hilbert space (also known as the reproducing kernel Hilbert space or RKHS), which is dense in $W$; let $\langle \cdot, \cdot \rangle_H$ denote the inner product on $H$. Then the inclusion map $i : H \to W$ is continuous, injective and has dense range, so the same is true of its adjoint $i^* : W^* \to H$. Let $W_* \subset H \subset W$ be the image of $i^*$.

Suppose $\{e_j\}$ is an orthonormal basis for $H$ which is contained in $W_*$, so that $e_j = i^* f_j$ for some $\{f_j\} \subset W^*$. Define the finite rank projection $P_n : W \to W_*$ by $P_n x = \sum_{j=1}^n f_j(x) e_j$.

Note that if $h \in H$ then $P_n h = \sum_{j=1}^n \langle e_j, h \rangle e_j$. So the restriction $P_n|_H$ is orthogonal projection and we have $P_n h \to h$ in $H$-norm. In other words, $P_n|_H \to I_H$ strongly on $H$, where $I_H$ is the identity map on $H$.

Moreover, it can be shown that $P_n \to I_W$ in $L^2(W, \mu; W)$. In other words, $\int_W \|P_n x - x\|_{W}^2 \mu(dx) \to 0$. In particular, there is a subsequence $P_{n_k}$ such that for $\mu$-almost every $x \in W$, we have $P_{n_k} x \to x$ in $W$-norm. (This is essentially the fact that $\|\cdot\|_W$ is a measurable norm on $H$ in the sense of Gross. I believe this result is due to Dudley, Feldman and Le Cam. A short proof due to Daniel Stroock can be found in these lecture notes by Bruce Driver; see Theorem 44.8.)

Now in some cases much more is true. For instance, consider the standard Wiener measure on $W = C_0((0,1])$ (i.e. all continuous paths starting at 0). Then $H$ is the Sobolev space $H^1_0((0,1])$. If we take $e_j$ to be the Schauder (aka Faber-Schauder) basis (i.e. the integrals of the Haar wavelets), then $P_n x$ is a piecewise linear function agreeing with $f$ at its vertices, which include all the dyadic rationals as $n \to \infty$. So $P_n x \to x$ uniformly, i.e. in $W$-norm, for every $x$ (not just almost every). In other words, $P_n \to I$ strongly on $W$.

As a related, not-quite example, let $W = \mathbb{R}^\infty$, viewed as the space of all real sequences with the product topology, and let $\mu$ be standard product Gaussian measure (I know this is not a Banach space). Then $H = \ell^2$. If we choose the standard basis $\{e_j\}$ for $H$, then $P_n x$ is merely the sequence that agrees with $x$ in its first $n$ terms and is 0 thereafter. Clearly $P_n x \to x$ in the $W$ topology, for every $x$.

I suppose this can't always happen, so I am looking for a counterexample.

What is an example of $W$, $\mu$ and $\{e_n\}$ such that it is not the case that $P_n \to I$ strongly on $W$? Better, is there an example where there is no subsequence $P_{n_k}$ with $P_{n_k} \to I$ strongly on $W$?

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    $\begingroup$ For an example, take for $W$ any separable Banach space that fails the approximation property. $\endgroup$ Commented Sep 27, 2015 at 16:04
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    $\begingroup$ @BillJohnson: Thanks. The failure of the approximation property says we cannot have $P_n \to I$ uniformly on compact sets; to see we also cannot have $P_n \to I$ pointwise, see here. To finish we need to know that such $W$ does admit a Gaussian measure, but it was shown by Gross that every separable Banach space does so (it's Remark 2 of his Berkeley symposium paper). Would you like to post an answer? $\endgroup$ Commented Sep 27, 2015 at 16:23

1 Answer 1


For an example, take for $W$ any separable Banach space that fails the approximation property, or even just does not have a Schauder basis. Every $\ell_p$ for $p\not= 2$ has such a subspace; see, for example, volume 1 of Classical Banach Spaces by Lindenstrauss and Tzafriri.


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