Dual space of $L^p(\Omega,\mathcal{A},\mu,\mathbb{R}^d)$.

Question

I want to show that for $p\in(1,+\infty)$ the dual space of $L^p(\Omega,\mathcal{A},\mu,\mathbb{R}^d)$ is isometrically isomorphic to $L^q(\Omega,\mathcal{A},\mu,\mathbb{R}^d)$, where $\frac{1}{p}+\frac{1}{q}=1$.

We have already shown the analogous result for $\mathbb{R}$ instead of $\mathbb{R}^n$ and therefore considered the bounded linear functional \begin{align*} \phi_g:f\mapsto\int_\Omega fg\,d\mu \end{align*} for every $g\in L^q(\Omega,\mathcal{A},\mu,\mathbb{R})$, where $f\in L^p(\Omega,\mathcal{A},\mu,\mathbb{R})$. We then proved that the mapping $g\mapsto\phi_g$ is an isometric isomorphism.

After several attempts, I have still not even figured out how to start, as I need to find an analogous functional $\phi_g$. I can not use the previous one with $fg$ considered to be a component-wise product, since the result would not be a scalar anymore, right? I have also thought about \begin{align*} \phi_g:f\mapsto\int_\Omega \langle f(x),g(x)\rangle\,d\mu(x), \end{align*} where $\langle.,.\rangle$ is the standard scalar product in $\mathbb{R}^d$. Is that an option?

Thanks in advance!

Edit: On $L^p(\Omega,\mathcal{A},\mu,\mathbb{R}^d)$ we define the norm \begin{align*} \|f\|:=\left(\int_\Omega\|f(x)\|_p^p\,d\mu(x)\right)^{\frac{1}{p}}, \end{align*} where $\|.\|_p$ is the $p$-norm on $\mathbb{R}^d$ and obtain a Banach space (I've already shown that).

Answer 1

You have this version of Hölder's Inequality (when using it with the discrete measure): $$ \left|\sum_j a_jb_j\right|\leq\left(\sum_j|a_j|^p\right)^{1/p}\left(\sum_j|b_j|^q\right)^{1/q}. $$ Then $$ \sum_{i=1}^d\left(\int|f_i|^p\right)^{\frac{1}{p}}\left(\int|g_i|^q\right)^{\frac{1}{q}} \leq \left(\sum_i \int|f_i|^p\right)^{1/p}\left(\sum_i \int|g_i|^q\right)^{1/q} =\left(\int\sum_i |f_i|^p\right)^{1/p}\left(\int\sum_i |g_i|^q\right)^{1/q} $$

Answer 2

Another way to see this is to use the following result: for Banach spaces $X_1,\dots,X_n,$ consider the direct sum, $$ X = \bigoplus_{i=1}^n X_i, $$ equipped with the $p$-norm, $$ \lVert (x_1,\dots,x_n) \rVert_{p} = \left(\lVert x_i\rVert_{X_i}^p\right)^{1/p}. $$ Then we can show that, $$ X^* \cong \bigoplus_{i=1}^n X_i^*, $$ equipped with the $p^*$-norm $\lVert \cdot \rVert_{p^*}$ defined similarly. Here the duality pairing is what you expect: $$ \left\langle (x_1,\dots,x_n),(f_1,\dots,f_n) \right\rangle = \sum_{i=1}^n \langle x_i, f_i \rangle. $$

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Now we can use the fact that, $$ L^p(\Omega,\mathcal{A},\mu,\mathbb R^d) \cong \oplus_{i=1}^n L^p(\Omega,\mathcal{A},\mu), $$ equipped with the $p$-norm as above. Hence the duality result shows that, $$ L^p(\Omega,\mathcal{A},\mu,\mathbb R^d)^* \cong \oplus_{i=1}^n L^p(\Omega,\mathcal{A},\mu)^* \cong \oplus_{i=1}^n L^{p^*}(\Omega,\mathcal{A},\mu) \cong L^{p^*}(\Omega,\mathcal{A},\mu,\mathbb R^d), $$ as required (the sums are equipped with the $p^*$ norms here). Note this gives the same pairing as the one you described, and both approaches (direct vs using the above result) boil down to the same argument.