# Solving the equation $Pu = f$, given that every $\ell \in \text{Ker}(P')$ has $\ell(f) = 0$

(Stanford Real Analysis Qualifying Exam: Spring 2012) (Ideal time: 24 minutes)

5) Let $X$, $Y$ be separable reflexive Banach spaces. Let $P \colon X \to Y$ be a bounded linear map, and $P'\colon Y^* \to X^*$ its adjoint (transpose). Suppose that there exists a Banach space $Z$ and a compact map $\iota\colon Y^* \to Z$ such that there exists $C > 0$ such that every $\phi \in Y^*$ satisfies $$\Vert \phi \Vert_{Y^*} \leq C(\Vert P'\phi \Vert_{X^*} + \Vert \iota\phi \Vert_Z).$$

b) Let $V \subset Y^*$ be a closed subspace with $V \oplus \text{Ker}(P') = Y^*$. Show that there is $C'>0$ such that every $\phi \in V$ has $\Vert \phi \Vert_{Y^*} \leq C'\Vert P'\phi\Vert_{X^*}$.

c) Show that if $f \in Y$ has $\ell(f) = 0$ for every $\ell \in \text{Ker}(P')$, then there exists $u \in X$ with $Pu = f$.

Note: Part (a) asks the student to show that $\text{Ker}(P')$ is finite-dimensional, which I've done.

My attempts:

(b) I think $P'|_V\colon V \to \text{Im}(P'|_V)$ is a bijection, and hence has a (bounded) inverse map $Q$. Therefore, if $\phi \in V$, then $\Vert \iota \phi\Vert_Z \leq \Vert \iota Q \Vert \Vert P'|_V \phi\Vert_Z$, and we take $C' = \Vert \iota Q\Vert$. However, I'm not fully confident in this solution.

(c) I see that every $\ell \in \text{Ker}(P')$ satisfies $\Vert \ell \Vert_{Y^*} \leq C \Vert \iota \ell\Vert_Z \leq c\Vert \ell\Vert_{Y^*}$ for some $c > 0$. This seems useful, but I'm not sure where to go from here.

Maybe we can find $u \in X$ via some sort of "weak limit" or as an element of $X^{**}$? I figure we should use the finite-dimensionality of $\text{Ker}(P')$ and/or the compactness of $\iota$ somehow.

• I think your attempt for (b) is correct: by taking a complement of $\ker P'$ and restricting $P'$ to its image (which is closed), you do create a bijection. However, I'm not sure why you need all the $\iota$'s in your proof though. For (c), assuming I'm interpreting the question correctly, can't you simply use the fact that $(\ker P')^\perp = \text{im} P$, which holds for all bounded linear operators $P$ with closed range?
– JHF
Mar 26 '13 at 16:53
• @JHF: Ah, yes, that should work for (c). Thanks. But why does $P$ have closed range? Is this because $V$ is closed? Also, if you write your comment as an answer, I'll accept it. Mar 26 '13 at 22:06
• @JHF: I don't know what you mean by all the $\iota$'s in my proof for (b), though. (Oh, and I should have said take $C' = \text{max}(C, C\Vert \iota Q\Vert)$.) Mar 26 '13 at 22:08

(b) To clarify what I meant about the $\iota$'s, I just thought it might be simpler to note that for $\phi \in V$, $\phi = QP'|_V \phi$, so $\lVert \phi \rVert_{Y^*} \leq \lVert Q \rVert \lVert P' \phi \rVert_{X^*}$, but maybe I'm missing something?
(c) We use the assumption $\lVert \phi \rVert_{Y^*} \leq C(\lVert P'\phi \rVert_{X^*} + \lVert \iota\phi \rVert_Z)$ to show that $P$ (or equivalently, $P'$) has closed range. (Indeed, this assumption is equivalent to having finite-dimensional kernel and closed range.)
First, recall that $P'|_V$ is injective by construction. Let $(u_n)$ be a sequence in $V$ such that $P'u_n \to x$ in $X^*$. We want to show that $x \in \text{im } P'$. To do this, note that $(u_n)$ is bounded; otherwise, let $v_n = \frac{u_n}{\lVert u_n \rVert_{Y^*}}$, so we would have $$\lVert v_n - v_m \rVert_{Y^*} \leq C(\lVert P' v_n - P' v_m \rVert_{X^*} + \lVert \iota v_n - \iota v_m \rVert_Z).$$
The right side converges to zero (along some subsequence) since $P' u_n$ has a limit, $(u_n)$ is unbounded, and $\iota$ is compact. So some subsequence of $(v_n)$ is Cauchy, hence has a limit $v$. However, we have $\lVert v \rVert_{Y^*} = 1$ but $P' v = 0$, contradicting injectivity.
Thus $(u_n)$ is bounded and therefore there is a convergent subsequence $(\iota u_n)$. Again using $$\lVert u_n - u_m \rVert_{Y^*} \leq C(\lVert P' u_n - P' u_m \rVert_{X^*} + \lVert \iota u_n - \iota u_m \rVert_Z) \to 0,$$ we see that $(u_n)$ has a Cauchy subsequence which converges to some $u \in Y^*$. Then $x = P'u$, so $\text{im } P'$ is closed as wanted.