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Section 5.5 Dual Spaces

In this section we provide several examples of dual spaces.

Subsection 5.5.1 Dual of a Hilbert Space

In a Hilbert space the dual can be identified with the original space. This is the key result of the Riesz Representation Theorem.
We need to prove that \(\rho\) is injective and surjective and that \(\|\rho(v)|_{H^*}=\|v\|_H\text{.}\)

First, let \(v,u\in H\) with \(v\neq u\) and assume that \(\rho(v)=\rho(u)\text{.}\) Then \(v-u\in H^\perp=\{0\}\text{,}\) so \(u=v\text{.}\)

Moreover, since \(|\langle v,w\rangle|\leq\|v\|\cdot\|w\|\text{,}\) we have that
\begin{equation*} \|\rho(v)\|_{H^*} = \sup_{w\neq0}\frac{|\langle v,w\rangle|}{\|w\|_H}\leq\|v\|_H. \end{equation*}
Since \(\frac{|\langle v,v\rangle|}{\|v\|_H}=\|v\|_H\text{,}\) it turns out that \(\|\rho(v)\|_{H^*}=\|v\|_H\text{.}\)

Finally, we show that \(\rho\) is surjective. Let \(\eta\in H^*\text{.}\) Then, by , there is a vector \(v\in H\) such that \(\ker\eta^\perp=\span\{v\}\text{.}\) We claim that there is a \(\lambda\) such that
\begin{equation*} \eta(w) = \langle\lambda v,w\rangle \end{equation*}
for all \(w\in H\text{.}\) Indeed, \(w=w_0\oplus\mu v\text{,}\) with \(w_0\in\ker\eta\text{,}\) so that
\begin{equation*} \eta(w) = \mu \eta(v) \end{equation*}
and
\begin{equation*} \langle\lambda v,w\rangle = \lambda\mu\|v\|^2_H. \end{equation*}
Then \(\lambda=\frac{\eta(v)}{\|v\|^2_H}\) does the job, namely
\begin{equation*} \eta(w) = \langle \frac{\eta(v)}{\|v\|^2_H}v,w\rangle \end{equation*}
for all \(w\in H\text{.}\) Equivalently,
\begin{equation*} \eta = \langle \frac{\eta(v)}{\|v\|^2_H}v,\cdot\rangle. \end{equation*}
Figure 5.5.2. Orthogonal decomposition on \(K\text{.}\)

Subsection 5.5.2 \((L^p(\Omega))^* \simeq L^q(\Omega)\) for \(\frac{1}{p}+\frac{1}{q}=1\)

Let \(\Omega\) be any open subset of a manifold. Then there is an invertable isometry between \((L^p(\Omega))^*\) and \(L^q(\Omega)\text{,}\) where \(q\) is the unique solution of \(\frac{1}{p}+\frac{1}{q}=1\text{,}\) for every \(p\in(1,\infty)\text{.}\) For a bounded smooth set \(\Omega\) and for \(1<p<\infty\) we use Hahn-Banach to show that \((L^p(\Omega))^*=L^q(\Omega)\text{,}\) where \(p\) and \(q\) are conjugate. For \(f\in L^q(\Omega)\) we define a linear functional through integration
\begin{equation*} \Phi_f(g) = \int_\Omega f(x) g(x) dx . \end{equation*}
By H\"olders inequality we have
\begin{equation*} |\Phi_f(g) |\leq \|f\|_q \|g\|_p. \end{equation*}
This shows that \(\Phi_f\in (L^p(\Omega))^*\) and that \(\|\Phi_f\|_{op}\leq \|f\|_q\text{.}\) But we can get more. We can show that we have an isometry between \(L^q\) and \((L^p)^*\text{.}\) To show this we choose \(g(x) = |f(x)|^{q-2} f(x) \text{.}\) Then
\begin{equation} \|g\|_p^p = \int_\Omega \left||f(x)^{q-2} f(x) \right|^p dx = \int_\Omega |f(x) |^q dx = \|f\|_q^q. \tag{5.5.1} \end{equation}
Hence \(\|g\|_p = \|f\|_q^{\frac{q}{p}}\text{.}\) Now
\begin{equation*} |\Phi_f(g) | = \int_\Omega f(x) |f(x)|^{q-2} f(x) dx = \int_\Omega |f(x)|^q dx = \|f\|_q^q = \|f\|_q \|f\|_q^{\frac{q}{p}} = \|f\|_q \|g\|_p. \end{equation*}
Hence
\begin{equation*} \|\Phi_f\|_{(L^p)^*} = \|\Phi_f\|_{op} = \|f\|_q, \end{equation*}
and the map \(L^q(\Omega) \to (L^p(\Omega))^*\) is an isometry.

What is missing is to show that each element in \((L^p)^*\) can be written as an integral operator. This is done via measure theory, and not covered in this book, but can be found in Halmos.

A simple proof can be done for \(p\in[1,2]\text{.}\) When \(p=2\text{,}\) it is enough using the Riesz Representation Theorem 5.5.1. Assume now that \(p\in[1,2)\) and assume, to make things simpler, that the measure of \(\Omega\) is 1. Then the identity map \(i:L^2(\Omega)\to L^p(\Omega)\) is a dense continuous embedding with \(\|f\|_{L^p(\Omega)}\leq\|f\|_{L^2(\Omega)}\text{.}\) Hence, by Proposition 5.3.5, \(i^*:(L^p(\Omega))^*\to (L^2(\Omega))^*\) is a continuous embedding and \(\|\eta\|_{(L^2(\Omega))^*}\leq\|\eta\|_{(L^p(\Omega))^*}\text{ for all }\eta\in(L^p(\Omega))^*.\) This means that every \(\eta\in(L^p(\Omega))^*\) is an element of \((L^2(\Omega))^*\) such that \(|\eta(f)|\leq K\|f\|_{L^p(\Omega)}\) for all \(f\in L^p(\Omega)\) and some \(K>0\text{.}\) As pointed out above, by the Riesz Representation we have that there is a \(g\in L^2(\Omega)\) such that
\begin{equation*} \eta(f) = \int_\Omega g(s)f(s)d\mu(s)\text{ for all }f\in L^2(\Omega). \end{equation*}
We claim that \(g\in L^q(\Omega)\text{.}\) To see this, we consider the sequence
\begin{equation*} f_k(x) = \min\{|g(x)|,k\}^{q-1}\sgn u(x). \end{equation*}
Notice that, for every \(k=1,2,\dots\text{,}\) \(f_k\in L^2(\Omega)\) and
\begin{equation*} \eta(f_k) = \int_\Omega \min\{|g(s)|,k\}^{q-1} |g(s)| d\mu(s) \geq \int_\Omega \min\{|g(s)|,k\}^{q}d\mu(s) = \|f_k\|^p_{L^p(\Omega)}. \end{equation*}
On the other side, by assumption,
\begin{equation*} K \|f_k\|^p_{L^p(\Omega)} \geq |\eta(f_k)|, \end{equation*}
so that
\begin{equation*} K \left(\int_\Omega \min\{|g(s)|,k\}^{q}d\mu(s)\right)^{1/p}\geq \int_\Omega \min\{|g(s)|,k\}^{q}d\mu(s). \end{equation*}
Since \(1/q=1-1/p\text{,}\) this means that
\begin{equation*} \|\min\{|g(s)|,k\}^{q}d\mu(s)\|_{L^q(\Omega)}\leq K\text{ for all }k=1,2,\dots, \end{equation*}
namely \(g\in L^q(\Omega)\text{.}\)

In any case, we have
\begin{equation*} (L^p(\Omega))^* \simeq L^q(\Omega) \end{equation*}
and we simply write
\begin{equation*} (L^p(\Omega))^* = L^q(\Omega). \end{equation*}
Because of the symmetry between \(p\) and \(q\text{,}\) this also shows that \(L^p\) spaces are reflexive for every \(p\in(1,\infty)\text{.}\)

Subsection 5.5.3 \((L^1(\Omega))^* \simeq L^\infty(\Omega)\) but \((L^\infty(\Omega))^* \supsetneq L^1(\Omega)\)

For \(f\in L^\infty(\Omega)\) we again define an integral operator as
\begin{equation*} \Phi_f(g) = \int_\Omega f(x) g(x) dx, \end{equation*}
with
\begin{equation*} |\Phi_f(g) | \leq \|f\|_\infty \|g\|_1. \end{equation*}
Similar as before we show that
\begin{equation*} \|\Phi_f\|_{(L^1)^*} = \|\Phi_f\|_{op} = \|f\|_\infty \end{equation*}
and we get an isometry
\begin{equation*} (L^1(\Omega))^* = L^\infty(\Omega). \end{equation*}
However, as we will not explicitly show here, we only have
\begin{equation*} L^1(\Omega) \subsetneqq (L^\infty )^* . \end{equation*}
The dual space \(M(\Omega) = (L^\infty(\Omega))^*\) is a measure space equipped with the total variation norm \cite{halmos}).

Subsection 5.5.4 \((L^p_{loc}(\bR))^* \simeq L^q_c(\bR)\) for \(\frac{1}{p}+\frac{1}{q}=1, p\in(1,\infty)\)

Recall that \(L^p_{loc}(\bR)\) is a Frechet space (see Subsection 3.6.2) and that \(f_n\to f\) in \(L^p_{loc}(\bR)\) if and only if \((f_n)|_{[-n,n]}\to f|_{[-n,n]}\) for every \(n=1,2,\dots\text{.}\)

Denote by
\begin{equation*} p_n(x) = \left(\int_{[-n,n]}|f(x)|^p dx\right)^{1/p} \end{equation*}
the seminorms giving rise to the topology of \(L^p_{loc}(\bR)\) and let \(\eta\in L^p_{loc}(\bR)^*\text{.}\) Then, by Corollary 4.2.15, there is a \(N>0\) such that
\begin{equation*} |\eta(f)| \leq M\cdot p_N(f)\text{ for all }f\in L^p_{loc}(\bR). \end{equation*}
In particular, if the essential support of \(f\) is disjoint from \([-N,N]\text{,}\) we have that \(\eta(f) = 0\text{.}\) Hence, there is a \(\hat\eta\in L^p([-N,N])^*\) such that
\begin{equation*} \eta(f) = \hat\eta(f)\text{ for all }f\in L^p_{loc}(\bR). \end{equation*}
Since \([-N,N]\) is compact, \(L^p([-N,N])^*\simeq L^q([-N,N])\) and so there is a \(g\in L^q([-N,N])\) such that
\begin{equation*} \eta(f) = \int_{[-N,N]}g(x)f(x)dx. \end{equation*}
We can extend \(g\) to a \(\tilde g\in L^q_c(\bR)\) by setting \(\tilde g=g\) within \([-N,N]\) and \(\tilde g=0\) otherwise, so that
\begin{equation*} \eta(f) = \int_{\bR}\tilde g(x)f(x)dx\text{ for all }f\in L^p_{loc}(\bR). \end{equation*}
Hence, the natural map \(J:L^q_c(\bR)\to L^p(\bR)^*\) is surjective. It is also injective since, given any \(h\in L^q_c(\bR)\text{,}\)
\begin{equation*} \int_{\bR}\tilde h(x)f(x)dx = 0\text{ for all }f\in L^p_{loc}(\bR) \end{equation*}
implies that \(h=0\text{.}\) Indeed, as we already did above, for \(\tilde h(x) = |h(x)|^{q-2}h(x)\text{,}\) we have that
\begin{equation*} \|\tilde h\|^p_{L^p(\bR)} = \|h\|^q_{L^q(\bR)}, \end{equation*}
so \(h\in L^p_{loc}(\bR)\) and
\begin{equation*} 0 = \int_\bR h(x)\cdot|h(x)|^{p-2}h(x)dx = \int_\bR |h(x)|^p dx = \|\tilde h\|^q_{L^q(\bR)}. \end{equation*}
The reader can show that this isomorphism is also an isometry. Finally, this result actually holds for any open set \(\Omega\subset\bR^n\text{,}\) namely
\begin{equation*} (L^p_{loc}(\Omega))^* \simeq L^q_c(\Omega)\text{ for }\frac{1}{p}+\frac{1}{q}=1, p\in(1,\infty) \end{equation*}

Subsection 5.5.5 \((L^p_{c}(\bR))^* \simeq L^q_{loc}(\bR)\) for \(\frac{1}{p}+\frac{1}{q}=1, p\in[1,\infty)\)

We did not discuss above about the topology of \(L^p_{c}(\bR)\text{.}\) As the dual of \(L^q_{loc}(\bR)\text{,}\) the topology on \(L^p_{c}(\bR)\) (see Definition 4.2.16) is induced by the family of seminorms
\begin{equation*} p_B(\eta) = \sup_{f\in B}|\eta(f)|, B\text{ bounded subset of }L^q_{loc}(\bR). \end{equation*}
What are bounded subsets of \(L^q_{loc}(\bR)?\) Recall that (Definition 4.2.7) a subset of a Frechet space is bounded if every seminorm is bounded on it. Applied to this case, it means that a set \(B\subset L^q_{loc}(\bR)\) is bounded if, for every compact \(K\subset\bR\text{,}\) the set \(B_K\subset L^q(K)\) of the restrictions to \(K\) of all functions in \(B\) is bounded in \(L^q(K)\) (that is a Banach space).

Recall \(\eta_n\to\eta\) in the strong "dual" topology means that \(p_B(\eta_n-\eta)\to0\) for every bounded subset \(B\) of \(L^q_{loc}(\bR)\text{.}\)

We claim that this topology coincides with the "natural" topology on \(L^q_{loc}(\bR)\text{,}\) namely the topology of \(L^q\) convergence over compact sets.

Indeed, assume that \(\eta_n\to\eta\) in the strong topology and let \(K\) be compact. Set
\begin{equation*} B_K = \{f\in L^p_c(\bR) : \supp(f)\subset K, \|f\|_{L^p(K)}\leq1\}. \end{equation*}
This is a bounded subset of \(L^p_{c}(\bR)\text{.}\) Hence
\begin{equation*} \|g_n-g\|_{L^p(K)} = \sup_{\|f\|_{L^p(K)}\leq1}\int_K f(x)(g_n(x)-g(x))dx = \sup_{f\in B_K}\int_K f(x)(g_n(x)-g(x))dx = p_{B_K}(g_n-g)\to0, \end{equation*}
namely \(g_n\to g\) in \(L^p(K)\text{.}\) Since this is true for every compact \(K\text{,}\) this means that strong convergence implies convergence over compacts. The reader can prove the inverse.

Now we show that the dual of \(L^p_{c}(\bR)\) is \(L^q_{loc}(\bR)\text{.}\) The fact that \(L^q_{loc}(\bR)\subset(L^p_{c}(\bR))^*\) comes from the elementary observation that, given any \(f\in L^q_{loc}(\bR)\text{,}\) then the map
\begin{equation*} \eta_f:g\mapsto\int_\bR f(x)g(x)dx \end{equation*}
is well-defined, linear and continuous. Continuity comes from the observation that, given a compact \(K\) and a sequence \(f_n\) converging to \(f\) in \(L^q_{loc}(\bR)\text{,}\)
\begin{equation*} \eta_f(g_n)-\eta_f(g)=\int_{K}(f_n(x)-f(x))g(x)dx \leq\|f_n-f\|_{L^p(K)}\|g\|_{L^p(K)}\to0. \end{equation*}
Hence, we just have to prove that all linear functionals on \(L^p_{c}(\bR)\) can be written this way. So, let \(\eta\in(L^p_{c}(\bR))^*\text{.}\) Since \(\eta\) is continuous, then its restriction to \(L^p([-n,n])\) is continuous and so there is an \(h_n\in L^q([-n,n])\) such that
\begin{equation*} \eta(f) = \int_\bR h_n(x)f(x)dx\text{ for all }f\text{ such that }\supp f\subset[-n,n].. \end{equation*}
The reader can verify that the \(h_n\) are compatible, in the sense that \(h_{n'}|_{[-n,n]}=h_n\) for \(n'>n\text{,}\) so this defines a function \(h\in L^q_{loc}(\bR)\) such that
\begin{equation*} \eta(f) = \int_\bR f(x)h(x)dx\text{ for all }f\in L^p_c(\bR). \end{equation*}

Finally, this result actually holds for any open set \(\Omega\subset\bR^n\text{,}\) namely
\begin{equation*} (L^p_{c}(\Omega))^* \simeq L^q_{loc}(\Omega)\text{ for }\frac{1}{p}+\frac{1}{q}=1, p\in[1,\infty) \end{equation*}

Subsection 5.5.6 \((C_c^\infty(\bR))^*=D^\infty(\bR) = \text{,}\) the space of Distributions

The elements of the dual of \(C_c^\infty(\bR)\) are called distributions. We denote their set by \(D^\infty(\bR)\text{.}\) The following are the most noteworthy types of distributions:

Regular distributions. There is a natural embedding of \(L^1_{loc}(\bR)\) into \(D^\infty(\bR)\) given by
\begin{equation*} g\mapsto \eta_g(f) = \int_\bR g(x)f(x)dx. \end{equation*}

Singular distributions. The most famous of these is Dirac's delta:
\begin{equation*} \delta_0(f) = f(0). \end{equation*}

Differentiation of distributions. One of the most important properties of distributions is that they can always be diffentiated infinitely many times. Indeed, over \(C_c^\infty(\bR)\) we have the operators
\begin{equation*} \frac{d^k}{dx^k}:C_c^\infty(\bR)\to C_c^\infty(\bR), k=1,2,\dots, \end{equation*}
and correspondingly on \(D^\infty(\bR)\) we have their adjoint
\begin{equation*} \left(\frac{d^k}{dx^k}\right)^*:D^\infty(\bR)\to D^\infty(\bR), k=1,2,\dots, \end{equation*}
so that
\begin{equation*} \left(\left(\frac{d^k}{dx^k}\right)^*\eta\right) (f) = \eta\left(\frac{d^k}{dx^k}f\right). \end{equation*}
It turns out that the most convenient way to define the derivative of a distribution is as follows:
\begin{equation*} \frac{d^k\eta}{dx^k}(f) = (-1)^k \left(\left(\frac{d^k}{dx^k}\right)^*\eta\right) (f) = (-1)^k \eta\left(\frac{d^k f}{dx^k}\right). \end{equation*}
With this definition, it is easy to check that, when \(g\in W^{1,1}(\bR)\text{,}\)
\begin{equation*} \frac{d}{dx}\eta_g = \eta_{\frac{d_w}{dx}g}, \end{equation*}
namely the weak and the distributional derivative. Notice that this example includes the \(C^1\) functions. On the other side, distributional derivatives always exist, whether weak do or not.
Consider the function \(g(x)=|x|\text{,}\) that has weak derivative equal to the step function \(s(x)=\begin{cases}\phantom{-}1,\amp x>0\\-1,\amp x<0\end{cases}\text{.}\) Then
\begin{equation*} \frac{d}{dx}\eta_g(f) = - \eta_g\left(\frac{d}{dx}f\right) = - \int_\bR g(x) f'(x) dx. \end{equation*}
Integrating by parts and using the fact that \(f\) has compact support, so that the finite part of the integration by parts is zero, we see that
\begin{equation*} \frac{d}{dx}\eta_g (f) = - \int_{(-\infty,0]} f(x) dx + \int_{[0,\infty)} f(x) dx = \int_\bR s(x)f(x)dx=\eta_s. \end{equation*}
How about the second derivative? The step function \(s\) has no weak derivative. If it had a derivative, then it would be zero everywhere except at most at \(x=0\) and so we would have \(\frac{d}{dx}\eta_s = \eta_{\frac{d_w}{dx}s}=0\text{,}\) but, given a \(\phi\in C_c^\infty(\bR)\text{,}\)
\begin{equation*} \frac{d}{dx}\eta_s(f) = \int_{(-\infty,0]} f'(x) dx - \int_{[0,\infty)} f'(x) dx = 2f(0), \end{equation*}
which is not necessarily zero. In fact, this shows that
\begin{equation*} \frac{d}{dx}\eta_s(f) = 2f(0) = 2\delta_0(f). \end{equation*}
Topology on \(C_c^\infty(\bR)\text{.}\) The Frechet topology on \(C^\infty(\bR)\) does not fit well with \(C_c^\infty(\bR)\text{.}\) On one side, \(C_c^\infty(\bR)\) is not complete in that topology. Moreover, the functional \(\eta_g\text{,}\) \(g\in L^1_{loc}(\bR)\text{,}\) is not necessarily continuous in that topology.

For instance, consider \(g(x)=1\text{.}\) If \(\eta_1\) were continuous then, given a sequence \(\phi_n\in C^\infty(\bR)\) such that \(\phi^{(k)}_n\to0\) uniformly on compact sets, it should happen that \(\eta_1(\phi_n)=\int_\bR\phi_n(x)dx\to0\text{.}\) It is easy, though, to build sequences for which this does not happen. For instance, let \(\phi(x)\) be a bump function with support in \([-1,1]\text{,}\) unitary \(C^0\) norm and integral equal to 1. Then the sequence \(\phi_n(x) = n\phi(x-n)\) converges to zero over every compact set but \(\eta_1(\phi_n)\to\infty\text{.}\) This phenomenon cannot take place if we ask an extra condition, namely that eventually the supports of all \(\phi_n\) are contained in a fixed compact set.

This topology is induced by the following family of seminorms:
\begin{equation*} p_{K,k}(f) = \|f\|_{C^k(K)}. \end{equation*}
Indeed, let \(f_n\to f\) and suppose that \(\cup\supp f_n\) is unbounded. Then there is a subsequence \(n_m\) and a sequence \(x_m\) such that \(x_m\in\supp f_{n_m}\) and \(|x_m|\to\infty\text{.}\) This is incompatible with the fact that the \(f_n\) converge to \(f\) uniformly over each compact set and that \(f\) has compact support.

Subsection 5.5.7 \((C^\infty(\bR))^*=D_c^\infty(\bR)\text{,}\) the space of Distributions with compact support

By definition, since \(C^\infty\) is metrizable, \(\eta\in(C^\infty(\bR))^*\) means that \(\eta(f_n)\to0\) for every \(f_n\to0\text{.}\)

We claim that this is equivalent to the fact that there is a compact set \(K\subset\bR\) such that
\begin{equation*} \eta(f) \leq C p_{K,k}(f)\text{ for all }f\in C^\infty(\bR),\;k=1,2,\dots. \end{equation*}
Indeed, suppose that there is no such \(K\text{.}\) Then, for every \(n=1,2,\dots\text{,}\) we can find a \(f_{n,k}\in C^\infty(\bR)\) such that
\begin{equation*} \eta(f_{n,k}) > n\cdot p_{K,k}(f_{n,k}). \end{equation*}
But then we can build a sequence \(\phi_{n,k}\to0\) with \(\phi(f_{n,k})\not\to0\text{,}\) contradicting the continuity of \(\eta\text{.}\)

Hence there is such \(K\) claimed above. We claim that \(\supp\eta\subset K\text{.}\) Indeed take a function \(\phi\in C^\infty(\bR)\) with \(\supp\phi\cap K=\emptyset\) and let \(\phi\in C^\infty(\bR)\) be a function with compact support such that \(\supp \psi\cap\supp\phi=\emptyset\text{.}\) Then
\begin{equation*} p_{K,k}(\psi) = 0 \end{equation*}
and so
\begin{equation*} \eta(\psi) \leq C p_{K,k}(f) = 0. \end{equation*}
Hence, \(\eta\) has compact support.

Subsection 5.5.8 Rainbows of functional spaces

Let \(M\) be a compact manifold endowed with a "Lebesgue class" measure -- this is a measure that, on every chart, differs from the Lebesgue measure by a positive \(L^1\) function. The reason for defining Lebesgue class measures is that in most manifolds there is no canonical measure and nevertheless any Lebesgue class measure gives rise to the same \(L^p\) spaces, so those spaces are canonical. Then the following sequences of functional spaces arises:
\begin{equation*} L^1(M)\supset\dots\supset L^p(M)\supset \dots\supset C^0(M)\supset C^1(M)\supset\dots\supset C^\infty(M) \end{equation*}
This sequence has the following important properties:
  1. each space in the sequence is complete with respect to its own standard topology;
  2. if \(X,Y\) appear in the sequence and \(X\subset Y\text{,}\) then \(X\) is dense in \(Y\text{.}\)
Because of ???, the respective duals are in a similar sequence with inverted order:
\begin{equation*} L^\infty_c(M)\subset\dots\subset L^q_c(M)\subset \dots\subset D^0_c(M)\subset D^1_c(M)\subset\dots\subset D^\infty_c(M) \end{equation*}
Unlike the original sequence, the inclusions are not necessarily dense. In fact, the only dense ones are \(L^p(M)\subset L^q(M)\) for \(p<q\text{.}\)

Assume now that \(M\) is not compact. Then we can build two sequences. One is
\begin{equation*} L^1_c(M)\subset\dots\subset L^p_c(M)\subset \dots\subset C^0_c(M)\supset C^1_c(M)\supset\dots\supset C^\infty_c(M) \end{equation*}
whose dual is given by
\begin{equation*} L^\infty_{loc}(M)\subset\dots\subset L^q_{loc}(M)\subset \dots\subset D^0(M)\subset D^1(M)\subset\dots\subset D^\infty(M). \end{equation*}
The other is
\begin{equation*} L^1_{loc}(M)\supset\dots\supset L^p_{loc}(M)\supset \dots\supset C^0(M)\supset C^1(M)\supset\dots\supset C^\infty(M), \end{equation*}
whose dual is given by
\begin{equation*} L^\infty_c(M)\subset\dots\subset L^q_c(M)\subset \dots\subset D^0_c(M)\subset D^1_c(M)\subset\dots\subset D^\infty_c(M). \end{equation*}