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I decompose $\mathbb{R}^n=\mathbb{R}^p\times \mathbb{R}^q$ for some $p,q\in \mathbb{N}$. Let $B\subseteq \mathcal{L}(\mathbb{R}^p,\mathbb{R}^q)$ be the closed unit ball with center $0$ (with respect to the operator norm). I define a map $$T:\mathcal{M}(B)\to \mathcal{M}(\mathbb{R}^n)$$ where $\mathcal{M}$ denotes the space of Radon measures as follows: Given $\mu\in \mathcal{M}(B)$, we define $T(\mu)$ by Riesz theorem as $$\int_{\mathbb{R}^n} f(v,w) T(\mu)(v,w)=\int_{B}\int_{\mathbb{R}^p}f(v,L(v))dm (v) d\mu(L) $$ where $f\in C_c(\mathbb{R}^n)$ and $dm$ is the Lebesgue measure.

Is the image of $T$ closed in the weak-topology?

(Does anyone know what the image of $T$ looks like or have anyone seen something like this in the literature)

(I think of $B$ as some chart of the Grassmannian manifold of $p$-dimensional subspaces in $\mathbb{R}^n$. My question can be changed by replacing $B$ with the Grassmannian manifold but the answer doesn't change from looking at local coordinates or looking globally)

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The image of $T$ is closed, where by "weak topology" I am assuming you mean the weakstar topology with respect to $C_c(\mathbb R^n)$. Your formula feels a little strange, since you're basically "integrating away" the variable $w$, but I'll assume that you wrote what you meant.

I claim that $F_f(L) := \int_{\mathbb R^p} f(v, Lv) ~dv$ is continuous for every $f \in C_c(\mathbb R^n)$. Indeed, if $L_j \to L$ then for every $v \in \mathbb R^p$, $(v, L_jv) \to (v, Lv)$, but since $f \in C_c(\mathbb R^n)$, $f(v, Lv) \leq 1_{B(0, R)} \|f\|_{C^0}$ where $R > 0$ is sufficiently large. So by dominated convergence, $\int_{\mathbb R^p} f(v, L_jv) ~dv \to \int_{\mathbb R^p} f(v, Lv) ~dv$ which is what we needed to show.

Now your formula reads $$\int_{\mathbb R^n} f(v, w) ~d(T\mu)(v, w) = \int_B F_f(L) ~d\mu(L)$$ so if $\mu_i \rightharpoonup^* \mu$ then (by continuity of $F_f$ and compactness of $B$) $$\int_{\mathbb R^n} f(v, w) ~d(T\mu_i)(v, w) \to \int_{\mathbb R^n} f(v, w) ~d(T\mu)(v, w).$$ In other words $T$ is continuous, with respect to the weakstar topology with respect to $C(B)$ and the weakstar topology with respect to $C_c(\mathbb R^n)$.

The only thing that can go wrong now is that $T$ fails to be proper. Suppose that $\mu_i(B) \to \infty$ and let $f(v, w) := 1_{B(0, 1)}(v) 1_{B(0, 100)}(w)$. Then for every $L \in B$, $$F_f(L) = \int_{B(0, 1)} 1_{B(0, 100)}(Lv) ~dv = \mathrm{vol}(B(0, 1)) =: \varpi$$ so $$\int_{\mathbb R^n} f(v, w) ~d(T\mu_i)(v, w) = \int_B F_f(L) ~d\mu_i(L) = \varpi \mu_i(B) \to +\infty.$$ Therefore $T\mu_i$ is has no convergent subsequence in $\mathcal M(\mathbb R^n)$ and therefore $T$ is proper, as desired.

Your construction vaguely reminds me of the decomposition of currents in Solomon's paper "X-rays of forms and currents" which is also about decomposing distributions along the Grassmannian. But it's not quite the same thing.

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  • $\begingroup$ Thank you very much for your answer and your reference. I wasn't aware of Solomon's paper. His construction is indeed very similar to mine. (To answer your remark that my formula looks strange. It is quite natural if you think of it as starting from $f\in C_c(\mathbb{R}^n)$ and then transforming $f$ to a function on the Grassmannian whose value at a linear subspace $L$ is $\int_L f$) $\endgroup$ Commented Jan 3 at 9:56
  • $\begingroup$ @OmarMohsen That makes sense! Somehow I missed that interpretation of your formula. $\endgroup$ Commented Jan 3 at 19:08
  • $\begingroup$ Actually, now that I think about that interpretation -- I think your construction is basically the adjoint of the X-Ray transform, which is the integral transform $f \mapsto (L \mapsto \int_L f)$. $\endgroup$ Commented Jan 3 at 19:18

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