Suppose $\prod^n L^1(I) $ is the product space of n $L^1$ integrable functions. What would be its dual space of continous linear functionals? would it be $\prod^n L^{\infty}(I)$? Do I need the norm for the product space to define the dual?
[Math] The dual space of a product space
functional-analysislebesgue-integralreal-analysis
Related Solutions
I have a super-naive interpretation that I find helpful. I think of the elements of $X$ as column vectors, and the elements of the dual, $X^*$ as row vectors. You can always multiply row vectors by column vectors, and each row vector gives a map from $X$ to $\mathbb{R}$.
In finite dimensions there is no canonical transformation of column vectors into row vectors, unless you choose a dot product. (If you haven't seen this before, then usual transpose operation corresponds to the usual dot product.) The same thing happens in infinite dimensions -- there is no canonical transformation unless you are in a Hilbert space. (The one thing that's different is that usually in infinite dimensions there's no transformation at all.)
The analogy is particularly clear for $L^p$ spaces. If you think of integration as fancy sums, then multiplying a row vector by a column vector in finite dimensions, $$ \sum_{i=1}^N v_i w_i, $$ becomes an integral for $f$ in $L^p$ and $g$ in its dual $L^q$, $$ \int f(x) g(x) dx. $$ The fact that they are all of this form is a non-trivial theorem, though, and there are elementary examples of function spaces where it fails.
Differentiation is more like a function from $X$ to $X$, so it's not a functional. You can define a functional by evaluating the derivative at a single point. Operators like differentiation on $L^p$ don't count not only because they are not continuous, but because they aren't defined on all of the space. This is usually the more important failure, and if you need differentiation for your application, you define a different function space that makes differentiation a continuous operator defined on that whole space. (Alternatively there's a weaker property than continuity, a "closed linear operator" and you can develop a theory for these.)
Your dual space $(X \times Y)^*$ is isometrically isomorphic to the product of dual spaces $X^* \times Y^*$ equipped with the norm $\|(f,g)\| = \sqrt{a\|f\|^2 + b\|g\|^2}$.
Proof:
Define $J : X^* \times Y^* \to (X \times Y)^*$ as
$$[J(f,g)](x,y) = af(x) + bg(y)$$
for $(f,g) \in X^* \times Y^*$.
$J$ is clearly well-defined and linear. Check that its inverse is given by
$$J^{-1}(F) = \left(\frac1a F(\cdot, 0), \frac1b F(0, \cdot)\right)$$
for $F \in (X \times Y)^*$ so $J$ is bijective.
We have
\begin{align} [J(f,g)](x,y) &= |af(x) + bg(y)| \\ &\le a|f(x)| + b|g(y)| \\ &\le a\|f\|\|x\| + b\|g\|\|y\| \\ &\le \sqrt{a\|f\|^2 + b\|g\|^2}\sqrt{a\|x\|^2 + b\|y\|^2} \\ &= \|(f,g)\|\|(x,y)\| \end{align}
so $\|J(f,g)\| \le \|(f,g)\|$.
To prove the converse inequality, let $\varepsilon > 0$ and pick unit vectors $x \in X$, $y \in Y$ such that $f(x) \ge (1-\varepsilon)\|f\|$ and $g(y) \ge (1-\varepsilon)\|g\|$.
Consider $\left(\|f\|x, \|g\|y\right) \in X \times Y$. Its norm is $$\left\|\left(\|f\|x, \|g\|y\right)\right\| = \sqrt{a\|f\|^2\|x\|^2 + b\|g\|^2\|y\|^2} = \sqrt{a\|f\|^2 + b\|g\|^2} = \|(f,g)\|$$
We have
\begin{align} J(f,g)(\|f\|x, \|g\|y) &= af(\|f\|x) + bg(\|g\|y) \\ &= a\|f\|f(x) + b\|g\|g(y) \\ &\ge a(1-\varepsilon)\|f\|^2 + b(1-\varepsilon)\|g\|^2 \\ &= (1-\varepsilon)(a\|f\|^2 + b\|g\|^2) \\ &= (1-\varepsilon)\|(f,g)\|^2 \\ &= (1-\varepsilon)\|(f,g)\|\|(\|f\|x, \|g\|y)\| \end{align}
so $\|J(f,g)\| \ge (1-\varepsilon)\|(f,g)\|$. Since $\varepsilon$ was arbitrary, we conclude $\|J(f,g)\| \ge \|(f,g)\|$.
Therefore $\|J(f,g)\| =\|(f,g)\|$ so $J$ is an isometric isomorphism.
This gives you an explicit formula for the dual norm of $F \in (X \times Y)^*$:
$$\|F\| = \sqrt{\frac1a\|F(\cdot, 0)\|^2 + \frac1b\|F(0, \cdot)\|^2}$$
Best Answer
Let $X$ be any Banach space, $n \in \mathbb{N}$ and $1 \le p \le \infty$.
On the space $\prod_{i=1}^n X$ we define the norm \begin{equation*} \lVert (x_1, \ldots, x_n) \rVert := \Big( \sum_{i = 1}^n \lVert x_i \rVert_X^p \Big)^{1/p}. \end{equation*} In case $p = \infty$ this is to be understood as \begin{equation*} \lVert (x_1, \ldots, x_n) \rVert := \max_{i = 1,\ldots,n} \lVert x_i \rVert_{X}. \end{equation*} Then, the dual of $\prod_{i=1}^n X$ is $\prod_{i=1}^n X^*$ with the norm \begin{equation*} \lVert (x_1^*, \ldots, x_n^*) \rVert := \Big( \sum_{i = 1}^n \lVert x_i^* \rVert_{X^*}^q \Big)^{1/q}, \end{equation*} where $q$ is the conjugate exponent to $p$, i.e., $1/p + 1/q = 1$. In case $q = \infty$, i.e., $p = 1$, this is to be understood as \begin{equation*} \lVert (x_1^*, \ldots, x_n^*) \rVert := \max_{i = 1,\ldots,n} \lVert x_i^* \rVert_{X^*}. \end{equation*}