I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
There is the following exercise in this book:
23 Suppose $V$ is finite-dimensional and $U$ and $W$ are subspaces of $V$. Prove that $(U \cap W)^0 = U^0 + W^0$.
$U^0$ is the set of linear functionals $\phi$ on $V$ such that $\phi(x) = 0$ for all $U$.
$W^0$ is the set of linear functionals $\phi$ on $V$ such that $\phi(x) = 0$ for all $W$.
$(U \cap W)^0$ is the set of linear functionals $\phi$ on $V$ such that $\phi(x) = 0$ for all $U \cap W$.
I think the assumption $\dim V < +\infty$ is not necessary.
But the author assumed that $\dim V < +\infty$.
Why?
My solution is here:
If $\phi \in (U \cap W)^0$, then $\phi(x) = 0$ for all $x \in U + W$.
Since $U \subset U + W$ and $W \subset U + W$, $\phi(x) = 0$ for all $x \in U \cap W$.
Conversely, if $\phi \in U^0 + W^0$, $\phi = \phi_1 + \phi_2$ for some $\phi_1 \in U^0$ and $\phi_2 \in W^0$.
For all $x \in U \cap W$, $\phi(x) = \phi_1(x) + \phi_2(x) = 0 + 0 = 0$.
So, $\phi \in (U \cap W)^0$.
Best Answer
Regarding your solution: it looks like you have not proven that $(U \cap W)^0 \subset U^0 + W^0$.
This is incorrect.
I'm not sure what you're trying to prove here, since we already have $\phi \in (U \cap W)^0$, which by definition means that $\phi(x) = 0$ for all $x \in U \cap W$.
This is correct. Indeed, we have $U^0 + W^0 \subset (U \cap W)^0$, even in the infinite-dimensional setting.