Tensor Product Structure of Sobolev Spaces and Generalization to $L^p$ Spaces

Question

I came across the following statement about Sobolev spaces:

$$ H^m(\Omega) \cong H^m(\Omega_1) \otimes \cdots \otimes H^m(\Omega_d) $$

where $ \Omega = \Omega_1 \times \cdots \times \Omega_d \subset \mathbb{R}^d$. I'd like to understand:

1. How exactly is this isomorphism defined?
2. Can this result be generalized to $ L^p $ spaces for $ p \neq 2 $, and if not, why?
3. Are there any beginner-friendly references on tensor products, especially for $ L^p (\Omega)$ spaces?

Any insights or precise references would be greatly appreciated!

PS: More precisely, Consider $L^p(\Omega)$, where $\Omega = [a_1, b_1] \times \dots \times [a_d, b_d]$ is a bounded rectangle in $\mathbb{R}^d$. For $\epsilon > 0$, does there exist a natural number $q$ such that, for any $f \in L^p(\Omega)$, there exist functions $f_{ij} : [a_j, b_j] \rightarrow \mathbb{R}$, with $i \in \{1, 2, \dots, q\}$ and $j \in \{1, \dots, d\}$, such that
$$ \Bigg|\Bigg| f - \sum_{i=1}^q \prod_{j=1}^d f_{ij}(x_j)\Bigg|\Bigg|_{L^p(\Omega)} \leq \epsilon? $$

Further, can this be done for any $p \in [0, \infty)$?
Also, can we choose $q$ depending only on $\epsilon, d, p$, such that the same $q$ works uniformly for any function $f \in L^p(\Omega)$?

Answer 1

The isomorphism is incorrect. Althought I do come across some similar Sobolev spaces which is denoted by $\widetilde H^m(\Omega)$. It would be good if you can provide more details on where you see the usage of such mixed norm Sobolev spaces.

What you ask for is something called Fubini property of the function spaces. For convenience I just state the case for $d=2$, since for large you just take more intersections / sums.

For $1<p<\infty$, $k\ge0$ and for bounded Lipschitz domains $\Omega_1,\Omega_2$ we have \begin{align*} W^{k,p}(\Omega_1\times\Omega_2)&=L^p(\Omega_1;W^{k,p}(\Omega_2))\cap L^p(\Omega_2;W^{k,p}(\Omega_1)); \\ W^{-k,p}(\Omega_1\times\Omega_2)&=L^p(\Omega_1;W^{-k,p}(\Omega_2))+L^p(\Omega_2;W^{-k,p}(\Omega_1)). \end{align*}

When I write papers I tend to use the language of strongly measurable vector-valued functions. When $p=1$, $L^1(\Omega_1;W^{k,1}(\Omega_2))=L^1(\Omega_1)\widehat\otimes_\pi W^{k,1}(\Omega_2)$ is the projective tensor.

In particular as tensor product of Hilbert spaces, \begin{align*} H^k(\Omega_1\times\Omega_2)&=L^2(\Omega_1)\widehat\otimes H^k(\Omega_2)\cap H^k(\Omega_1)\widehat\otimes L^2(\Omega_2); \\ H^{-k}(\Omega_1\times\Omega_2)&=L^2(\Omega_1)\widehat\otimes H^{-k}(\Omega_2)+H^{-k}(\Omega_1)\widehat\otimes L^2(\Omega_2). \end{align*}

When $\Omega_j=\mathbb R^{n_j}$ is the total spaces and $k\ge0$, see e.g. Hans Triebel, Theory Of Function Spaces Chapter 2.5.13. I don't have a good reference for the general case. I can write the proof down (or shameless cite my own paper) if you need.

Update: Anyway let me list a couple definitions here first. I have to say that I am not an expert as well, so there may be better references.

Let $(\Omega,\Sigma,\mu)$ be a measure space and let $X$ be a Banach space. We say a map $f:\Omega\to X$ is a simple function, if it is of the form $f=\sum_{j=1}^N\chi_{E_j}\otimes x_j$ in the senses that $f(\omega)=\sum_j\chi_{E_j}(\omega)\cdot x_j$, where $E_j\in\Sigma$ are measurable sets and $x_j\in X$.

Here we can define integral $\int_\Omega\sum_{j=1}^N\chi_{E_j}\otimes x_j d\mu=\sum_{j=1}^N\mu(E_j)\cdot x_j\in X$.

We say $f:\Omega\to X$ is strongly measurable, if it is a pointwise limit of a sequence of simple functions.

For $1\le p\le\infty$, the Bochner space $L^p(\mu;X)$ is the space of all stronly measurable $f$ such that $\|f\|_{L^p(\mu;X)}=(\int_\Omega\|f(\omega)\|_X^pd\mu(\omega))^{1/p}<\infty$, quotient of $\mu$ almost everywhere zero functions. Here in the case $p=\infty$ we use $\|f\|_{L^\infty(\mu;X)}=\inf\{r>0:\mu\{\omega:\|f(\omega)\|>r\}=0\}$.

The Fubini theorem states that for $1\le p<\infty$, $L^p(\Omega_1\times\Omega_2)=L^p(\Omega_1;L^p(\Omega_2))=L^p(\Omega_2;L^p(\Omega_1))$. See for example Analysis in Banach spaces. Vol. I. Section 1.2. However this is not true for $p=\infty$.

In the case $p=\infty$ we have $L^\infty(\Omega_1;L^\infty(\Omega_2))\subsetneq L^\infty(\Omega_1\times\Omega_2)$. A typical example is $\chi_{\mathbb R_+}(x-y)$ for $x,y\in\mathbb R$.

If you really need to work with $p=\infty$, then one should use $L^\infty(\Omega_1\times\Omega_2)=\mathscr L(L^1(\Omega_1);L^\infty(\Omega_2))$, in the sense we identify $K(x,y)\in L^\infty_{x,y}$ with $T^K:L^1(\Omega_1)\to L^\infty(\Omega_2)$ by $T^Kg(y)=\int_{\Omega_1}K(x,y)g(x)dx$. See Delio Mugnolo & Robin Nittka (2010) Properties of representations of operators acting between spaces of vector-valued functions for a general servey.

We can define vector-valued Sobolev spaces $W^{k,p}(\Omega;X)$ in a similar way. I refer to Analysis in Banach spaces. Vol. I. Section 2.5.

If you use the following norm definition $\|f\|_{W^{k,p}(\Omega;X)}^p=\sum_{|\alpha|\le k}\|\partial^\alpha f\|_{L^p(\Omega;X)}^p$. Then for $1\le p\le\infty$ and $k\ge0$, \begin{equation*} W^{k,p}(\Omega_1\times\Omega_2)=\bigcap_{j=0}^kW^{j,p}(\Omega_1;W^{k-j,p}(\Omega_2))\text{ with }\|f\|_{W^{k,p}(\Omega_1\times\Omega_2)}^p=\sum_{j=0}^k\|f\|_{W^{j,p}(\Omega_1;W^{k-j,p}(\Omega_2))}^p. \end{equation*}

I don't have a good reference for the above equality and I proved it myself in an ongoing paper.

Next let's turn to tensor products. On vector spaces it is a bad idea to refer to universal property. Instead one should think of the idea that tensor of two vectors becomes a matrix.

Given two Banach spaces $X$ and $Y$, the projective tensor space $X\widehat\otimes_\pi Y$ is the closure of the algebra tensor space $X\otimes Y$ under the norm $$\|f\|_{X\widehat\otimes_\pi Y}=\inf\Big\{\sum_{j=1}^\infty\|x_j\|_X\|y_j\|_Y:f=\sum_{j=1}^\infty x_j\otimes y_j\Big\}.$$ The projective tensor can be defined by the universal property via the category of Banach spaces.

However there is a different space called injective tensor space $X\widehat\otimes_\epsilon Y$, which is the closure under the norm $$\|f\|_{X\widehat\otimes_\epsilon Y}=\sup\Big\{\sum_{j=1}^\infty \phi(x_j)\psi(y_j):f=\sum_{j=1}^\infty x_j\otimes y_j,\quad\phi\in X',\psi\in Y',\ \|\phi\|_{X'}=\|\psi\|_{Y'}=1\Big\}.$$

We always have $X\widehat\otimes_\pi Y\subset X\widehat\otimes_\epsilon Y$. For infinite dim Banach spaces they are never equal. In most cases (say $X$ and $Y$ reflexive) we have $(X\widehat\otimes_\pi Y)'=X'\widehat\otimes_\epsilon Y'$.

Answer 2

For $L^p$ spaces with $p \in [1,\infty)$ specifically, it is almost correct.

For two vector spaces $V, W$, the tensor product $V\otimes W$ is characterized by this theorem. (There are other definitions; note that the definitions using (Hamel) bases is not super useful when working with $L^p$ spaces, which come more naturally with Schauder bases.)

Take $V = L^p(\Omega_1)$ and $W = L^p(\Omega_2)$, and define the bilinear injective operation $$\phi: (v,w) \mapsto ((x,y) \in \Omega_1\times \Omega_2 \mapsto v(x) w(y)) . $$ Clearly, if $(v,w)\in V\times W$ then $\phi(v,w) \in L^p(\Omega_1 \times \Omega_2)$ by Fubini's theorem. To apply the theorem above, we need to check that

1. the image of $\phi$ spans $L^p(\Omega_1 \times \Omega_2)$,
2. $V, W$ are $\phi$-linearly disjoint.

The first one is almost true, in that the span of the image of $\phi$ is dense in $L^p(\Omega_1\times \Omega_2)$.

Proof: This follows from classical ideas in measure theory. Simple functions are dense in $L^p(\Omega)$ for every Borel-measurable subset $\Omega$ of $\mathbb R^d$; by outer regularity of the Lebesgue measure, any $\mathbf 1_A$ with $A$ a Borel set of $\Omega$ can be well-approximated in $L^p$ norm by $\mathbf 1_O$ with $O$ an open set of $\Omega$; and any open set of $\Omega_1 \times \Omega_2$ can be well-approximated by a finite disjoint union of open sets of the form $(I_1 \cap \Omega_1) \times (I_2 \cap \Omega_2)$ with $I_1$, resp. $I_2$ boxes of $\mathbb R^{d_1}$, resp. $\mathbb R^{d_2}$, with rational coordinates. Combining everything together, we can approximate any function in $L^p(\Omega_1 \times \Omega_2)$ by a finite sum of $\phi(v,w)$ with $v = \mathbf 1_{\{I_1 \cap \Omega_1\}}$ and $w = \mathbf 1_{\{I_2 \cap \Omega_2\}}$. We deduce $\overline{\mathrm{span}(\phi(V \times W))} = L^p(\Omega_1 \times \Omega_2)$, where the completion is with respect to the $L^p$ norm. $\square$

This is where it becomes clear that the tensor product construction is limited, and that you need to take completeness into account. The topological tensor product, which @Liding Yao mentioned in their answer (a particular case is the projective tensor $\hat\otimes_\pi$), completes the usual tensor product (for a certain "cross-norm" $\pi$) to obtain a complete space.

So, if we do not have $L^p(\Omega_1 \times \Omega_2) = V \otimes W$, can we maybe show that $L^p(\Omega_1 \times \Omega_2)$ is the completion of $V \otimes W$ for the $L^p$ norm? Using another definition of the tensor product, namely the universal property, we obtain from $\phi$ an injection $i : V \otimes W \to L^p(\Omega_1 \times \Omega_2)$, and we know that $\overline{\mathrm{span}(i(V\otimes W))} = L^p(\Omega_1 \times \Omega_2)$, where the completion is with respect to the $L^p$ norm. In that sense, we have $$ L^p(\Omega_1 \times \Omega_2) = L^p(\Omega_1) \hat\otimes_{\|\cdot\|_p} L^p(\Omega_2) = \overline{L^p(\Omega_1) \otimes L^p(\Omega_2)} .$$ In fact, we can check that if $p=1$, we can replace $\hat\otimes_{\|\cdot\|_1}$ by $\hat\otimes_\pi$, by observing that $\pi(\sum_{i=1}^n \mathbf{1}_{A_i}) = \|\sum_{i=1}^n \mathbf{1}_{A_i}\|_1$ when $A_i = I_i \cap (\Omega_1 \cap \Omega_2)$ with $(I_i)_i$ disjoint boxes. This gives our final conclusion: $$ L^1(\Omega_1 \times \Omega_2) = L^1(\Omega_1) \hat\otimes_\pi L^1(\Omega_2) . $$

Answer 3

My (affirmative) answer concerns the Hilbert tensor product (as defined e.g. here) of the Hilbert spaces $H^k(\Omega)$ for sufficiently nice (e.g. Lipschitz) bounded domains $\Omega_j$ for all $j\leq d$. I would imagine that the case of nice unbounded domains may be treated similarly (because of the heuristics below) but giving an explicit isomorphism is more involved (because the spectra below will not be discrete).

In the case of spaces $H^1$ ($m=1$), a detailed proof for bounded domains with additional properties (slightly more general than requiring the $\Omega_j$'s to be Lipschitz) can be found here (Cor. 6.3). The strategy is similar to the one adopted below, and more explicit.

My definition of choice is the one of Bessel potential space: \begin{gather} H^k(\Omega):=\{f\in L^2(\Omega) : (1-\Delta_\Omega)^{k/2}f\in L^2(\Omega)\}, \\ \lVert f\rVert_{H^k(\Omega)}:= \lVert(1-\Delta_\Omega)^{-k/2} f\rVert_{L^2(\Omega)}. \end{gather} (Here, $\Delta_\Omega$ is the Laplacian on $\Omega$ with Neumann boundary conditions, i.e. the one with form domain $H^1(\Omega)$. This definition is equivalent to the more standard ones, on nice domains and on manifolds, see e.g. here (§4).) When $k$ is odd, $(1-\Delta_\Omega)^{k/2}$ is the square-root operator of the non-negative self-adjoint operator $(1-\Delta_\Omega)^k$, constructed (for instance) by means of the continuous functional calculus for self-adjoint operators. In all cases $(1-\Delta_\Omega)^{-k/2}$ is the inverse of $(1-\Delta_\Omega)^{k/2}$. When $k$ is even, it is not difficult to show that it is a power of the resolvent of $-\Delta_\Omega$ computed at $1$. The same holds for all real $k$. (See again here (§4) or here.)

Note. With the definition of fractional powers of $1-\Delta_\Omega$ via the heat kernel (see here) it is possible to extend the arguments below to the case of real $k$.

The heuristics is very simple:

Heuristics. $(1-\Delta_\Omega)^{-k/2}\colon H^k(\Omega)\to L^2(\Omega)$ is a unitary operator. (The same applies to $\Omega_j$ for every $j\leq d$.) Then the conclusion, since the tensor product of unitary operators is unitary and $$L^2(\Omega)\cong \bigotimes_j^d L^2(\Omega_j).$$ (This latter statement I would like to take for granted throughout.)

Now let's move to a proof sketch. Denote by $\sigma(A)$ the spectrum of an operator $A$.

Fact 1. We have $$\sigma(1-\Delta_\Omega)=\sigma(1-\Delta_{\Omega_1})+\cdots+\sigma(1-\Delta_{\Omega_d}).$$

! Proof. Note that $\Delta_\Omega = \bigoplus_j^d \Delta_{\Omega_j}$. For the Laplacian(s) with Neumann boundary condition this holds (for instance) because on a product domain we can factor the normal derivatives at the boundary (which is a product of boundaries times domains). The conclusion follows from the analogous statement for the heat-semigroup operators $H_{\Omega_j,t}:=e^{-t(-\Delta_{\Omega_j})}$ (the advantage is that these are bounded operators and more standard facts apply), and since the spectra $\sigma(\Delta_j)$ are discrete by boundedness of $\Omega$ and in one-to-one correspondence with $\sigma(H_{\Omega_j,t})$ via $t\mapsto e^{-t}$.

In fact, a much stronger statement is true.

Fact 2. The eigenspace decomposition of $(1-\Delta_\Omega)^{k/2}$ splits over the tensor product. That is $$L^2(\Omega) = \bigoplus_i E_{\lambda_{\Omega,i}} = \bigoplus_{i_1,\dotsc, i_d} \bigotimes_j^d E_{\lambda_{\Omega_j,i_j}}= \bigotimes_j^d L^2(\Omega_j).$$ where $E_{\lambda_{\Omega_j,i}}$ is the eigenspace in $L^2(\Omega_i)$ of the $i$-th eigenvalue $\lambda_{\Omega_j,i}$ for $(1-\Delta_{\Omega_j})^{k/2}$.

! Proof. It is clear (cf. here, Thm. 3.2) that if $\psi_{n,j}$ is an eigenfunction of $(1-\Delta_{\Omega_j})^{k/2}$ in an $L^2(\Omega_j)$-orthonormal bases for the $i$-th eigenspace $E_{\lambda_{\Omega_j,i}}$ of $\Delta_{\Omega_j}$, then $$\bigotimes_j^d \psi_{n_j, j} \quad \text{is an eigenfunction for $(1-\Delta_\Omega)^{k/2}$ in the corresponding eigenspace.}$$ (Here, the index $i$ counts eigenvalues without multiplicity, while the index $n$ counts eigenvalues with multiplicity.)
Indeed, for $k$ even the above assertion follows from the decomposition of $\Delta_\Omega$ in the proof of Fact 1. For $k$ odd it holds since the same decomposition commutes with taking square roots of the operators because the spectrum of all operators involved is discrete.
This shows the $\supset$ inclusion. (For the case of $H^1$ cf. here, Thm. 6.2.)
The reverse inclusion follows if we show that the system of eigenfunctions given above in the form of elementary tensors is complete in $L^2(\Omega)$. This must be the case since $\bigotimes_j^d L^2(\Omega_j)=L^2(\Omega)$ and for each $j\leq d$ the system $\{\psi_{n_j,j}\}_{n_j}$ is a complete orthonormal system in $L^2(\Omega_j)$.

Now, we can give the isomorphism explicitly: $$ J : f=\sum_n \langle f | \psi_n\rangle_{H^k(\Omega)} \psi_n \longmapsto \sum_{n_1,\dotsc, n_j}\langle f | \psi_n\rangle_{H^k(\Omega)} \bigotimes_j^d \psi_{n_j,j}, \qquad \text{with} \quad \psi_n= \bigotimes_j^d \psi_{n_j,j}.$$

Answer 4

After reading all the above comments this is what I thought should answer my questions. Please correct me if I am wrong. (Thanks in advance!)

1. Approximation by a simple function: Let $f \in L^p(\Omega)$, where $\Omega = [a_1, b_1] \times \dots \times [a_d, b_d] \subset \mathbb{R}^d$ is a bounded rectangle. Since any measurable function can be approximated in $L^p$-norm by a simple function, there exists a simple function $s(x) = \sum_{k=1}^m c_k \chi_{E_k}(x)$ such that $\|f - s\|_{L^p(\Omega)} \leq \epsilon/2$, where each $E_k \subset \Omega$ is a measurable set, and $c_k \in \mathbb{R}$ are constants.

2. Approximating the simple function using products of one-dimensional functions: Each measurable set $E_k$ can be approximated in measure by a finite union of rectangles $\{R_{k,l}\}_{l=1}^{n_k}$ such that $|E_k \triangle (\cup_{l=1}^{n_k} R_{k,l})| \leq \delta$, with $\triangle$ denoting the symmetric difference. By choosing $\delta$ sufficiently small, we ensure that replacing $E_k$ with these rectangles introduces an error in the $L^p$-norm of at most $\epsilon/2$. Therefore, we can approximate $s(x)$ by a function of the form $$ \tilde{s}(x) = \sum_{k=1}^m c_k \sum_{l=1}^{n_k} \chi_{R_{k,l}}(x), $$ with $\|s - \tilde{s}\|_{L^p(\Omega)} \leq \epsilon/2$.

3. Expressing characteristic functions of rectangles in product form: Each rectangle $R_{k,l} = \prod_{j=1}^d [\alpha_{k,l}^j, \beta_{k,l}^j]$ can be written as a product of one-dimensional characteristic functions: $$ \chi_{R_{k,l}}(x) = \prod_{j=1}^d \chi_{[\alpha_{k,l}^j, \beta_{k,l}^j]}(x_j). $$ Hence, $\tilde{s}(x)$ becomes a finite sum of products of one-dimensional functions.

Conclusion Combining the above steps, we approximate $f$ in $L^p$-norm by a finite sum of products of one-dimensional functions, with the total error not exceeding $\epsilon$. The number of product terms $q$ depends on $\epsilon, d, p$, and $f$. While $q$ can be chosen depending only on $\epsilon, d, p$, and $f$, it is unlikely that $q$ can be made independent of $f$ in general.The argument should work for $1\leq p < \infty$

Answer 5

A very useful tool which can be used to approach problems of this kind is the Stone-Weierstrass approximation theorem, the proof can be found in Principles of Mathematical Analysis - Walter Rudin, I state it here.

Suppose $X$ is a compact topological space and $\mathcal{A}$ is a subset of $C(X,\mathbb{R})$ satisfying the following properties:

1. $f,g \in \mathcal{A},a,b \in \mathbb{R} \implies a f + b g \in \mathcal{A}$
2. $f,g \in \mathcal{A} \implies f \cdot g \in \mathcal{A}$
3. $\forall x \in X \; \exists f \in \mathcal{A} \; : \; f(x) \neq 0$
4. $\forall x,y \in X \; \exists f \in \mathcal{A} \; : \; f(x) \neq f(y)$ Then taking the closure with respect to the uniform convergence topology we have $\overline{\mathcal{A}} = C(X,\mathbb{R})$, which means that every continuos function on $X$ can be approximated uniformly with functions inside $\mathcal{A}$.

In the case of $X = [0,1]$, $\mathcal{A} = \mathbb{R}[x]$ this gives the classical Weierstrass approximation theorems, moreover this can be used for $X = \mathbb{R}/(2\pi \mathbb{Z})$ and $\mathcal{A} = span(\{1,\sin(nx),\cos(nx) \; : \; n \geq 1\})$ to show that every $2\pi$-period continuos function can be approximated uniformly by trigonometric polynomials. I show how this theorem can be used to show that $L^p(\Omega_1 \times \Omega_2) \cong L^p(\Omega_1) \otimes L^p(\Omega_2)$ in the case of $\Omega_1,\Omega_2$ open subsets in $\mathbb{R}^{d}$. Consider $\mathcal{A} := span\{ f(x,y) = \alpha(x)\cdot \beta(y) \; : \; \alpha \in C^\infty_c(\Omega_1)\, , \, \beta \in C^\infty_c(\Omega_2)\}$, clearly $\mathcal{A}$ satisfies all the conditions of the previous theorem, $X = \Omega_1 \times \Omega_2$ is not compact but we can make slightly modify the theorem to show that $C_c(\Omega_1 \times \Omega_2) = \overline{\mathcal{A}}$, then using the density of $C_c(\Omega_1 \times \Omega_2)$ in $L^p(\Omega_1 \times \Omega_2)$ and the trivial inequality $\|f\|_{L^p(\Omega)} \leq \|f\|_{L^\infty(\Omega)}|supp(f)|^{1/p}$ we have that $\mathcal{A}$ is dense in $L^p(\Omega_1 \times \Omega_2)$, but since clearly $\mathcal{A} \subset L^p(\Omega_1) \otimes L^p(\Omega_2)$ we also have $L^p(\Omega_1 \times \Omega_2) \cong L^p(\Omega_1) \otimes L^p(\Omega_2)$.\

The Stone-Weierstrass theorem works well with the uniform norm, if we need to use some norms involing derivatives it's not well suited anymore but sometimes there are some tricks to make it work, I provide an example.

I show that $\mathbb{R}[x,y]$ is dense in $C^1([0,1]\times[0,1])$, using Stone-Weierstrass theorem we know that $\mathbb{R}[x,y]$ is dense in $C([0,1] \times [0,1])$, now we can decompose $f \in C^1([0,1] \times [0,1])$ in the following way:

$$f(x,y) = f(0,0) + \int_{0}^{x}\partial_x f(t,0)dt + \int_{0}^{y} \partial_y f(x,t)dt$$, now using the previous result for every $\epsilon > 0$ we can find $p^{(\epsilon)}_1(x,y),p^{(\epsilon)}_2)(x,y) \in \mathbb{R}[x,y]$ such that $$\|\partial_x f - p^{(\epsilon)}_1\|_{\infty} + \|\partial_y f - p^{(\epsilon)}_2\|_{\infty} < \epsilon$$, so I can define $$p^{(\epsilon)}(x,y) := f(0,0) +\int_{0}^{x} p^{(\epsilon)}_1(t,y)dt + \int_{0}^{y} p^{(\epsilon)}_2(x,t)dt $$ and clearly $p^{(\epsilon)}$ is in $\mathbb{R}[x,y]$ and we can prove that $\|f - p^{(\epsilon)}\|_{C^1([0,1] \times [0,1]} < \epsilon$. There is a reason why I showed this example, we can use this to prove the initial claim, in the case of $\Omega_i$ bounded with $C^1$-boundary there exists an extension operator $E \; : \; H^m(\prod_{i}\Omega_i) \to H^m(Q)$ where $Q$ is a really big cube, now using the previous example we have that the polynomials are dense in $C^m(Q)$, since this norm is stronger then the $H^m(Q)$ norm the polynomials are dense also in $H^m(Q)$,so this implies that $\otimes_{i} H^m(\Omega_i)$ is dense in $H^m(Q)$, but then we can consider the restriction operator $R \; : \; H^m(Q) \to H^m(\prod_i \Omega_i)$, and since clearly $R$ maps every polynomial in $\otimes_i H^m( \Omega_i)$ we have that $\otimes_i H^m(\Omega_i)$ is dense in $H^m(\Omega)$ and this shows that $\otimes_i H^m(\Omega_i) \cong H^m(\Omega)$