Let $D:X\rightarrow Y$ be a bound linear operator.
$\newcommand{\im}{\operatorname{im}}$
The claim given is:
A closed image $D(X)$ and finite dimensional cokernel is equivalent to a closed image $D^*(Y^*)$ of the transpose and finite dimensional kernel.
I am stuck on my way to show the properties for the transposed function $D^*$. I have some notes from class, that is what I was able to complete. I showed that if an element $a$ of the kernel is $0$ for all basis vectors of the cokernel, then it is also $0$. Thus $\dim(\ker(D^*))\leq \dim(\operatorname{coker}(D))<\infty$.
Now let $\{\beta_n\}$ be a sequence such that $D^*\beta_n\rightarrow \alpha\in X^*$. I want to show that $\im(D^*)$ is closed so in this case that $\alpha\in \im(D^*)$. Im my notes there is just the remark that we can deduce the rest from the fact that $\im(D)$ is closed, but I don't see how.
Best Answer
I'm assuming $X$ and $Y$ are normed vector spaces and that $X^*$ and $Y^*$ are equipped with the standard dual norm, i.e. the norm of $\phi$ in $X^*$ is $\underset{v \neq 0}{\mathrm{sup}} \frac{|\phi(v)|}{\|v\|_X}$. I haven't done functional analysis for a long time, so I might have over-detailed things, but better safe than sorry!
($\Rightarrow$) Suppose $D(X)$ is closed and $Y/D(X)$ is finite-dimensional. Let $\{\phi_n\}$ be a sequence in $Y^*$ such that $\{D^*(\phi_n)\}$ converges to $\psi$ in $X^*$.
We can re-write $D^*(\phi_n)(x)$ as $\phi_n(D(x))$, and since $Y/D(X)$ is finite-dimensional, we can find a basis of $Y/D(X)$, say $\{y_1+D(X), \cdots, y_n+D(X) \}$ such that any vector in $Y$ has a unique writing of the form $y = \sum_{i=1}^n a_i y_i + d$, and $y$ is expressed uniquely as a function of $a_1,\cdots,a_n$ in $\mathbb R$ (or $\mathbb C$, don't know which field you're working over) and $d \in D(X)$. For $\phi \in Y^*$, this means that $\phi$ is uniquely determined by $\phi(y_1),\cdots,\phi(y_n)$ and $\phi|_{D(X)}$. In particular, if $D^*(\phi)$ is zero, then $\phi|_{D(X)} = 0$, so that $\ker D^*$ is a vector space of dimension $n$ spanned by the $n$ functionals $\omega_1,\cdots,\omega_n$ such that $\omega_i(y_j) = \delta_{ij}$ and $\omega_i|_{D(X)} = 0$ for $i,j=1,\cdots,n$.
The information that $\{D^*(\phi_n)\}$ converges to $\psi \in X^*$ tells us that $\|\phi_n \circ D - \psi\|_{X^*} \to 0$, or in other words, that $\underset{x \neq 0}{\mathrm{sup}} \frac{|\phi_n(D(x)) - \psi(x)|}{\|x\|_X} \to 0$, and in particular $\phi_n(D(x)) \to \psi(x)$. It follows that for any $d \in D(X)$, $\phi_n(d)$ is a convergent sequence. Denote by $\phi(d)$ its limit, so we get a function $\phi: D(X) \to \mathbb R$, which we extend by zero to all of $Y$, i.e. $\phi\left(\sum_{i=1}^n a_i y_i + d\right) \overset{\mathrm{def}}{=} \phi(d)$.
We wish to show that $\phi \in Y^*$ and that $D^*(\phi) = \psi$. The latter is by construction. As to the former, the linearity of $\phi$ follows from the linearity of $\psi$. For continuity, suppose a sequence $d_k$ converges to $d \in D(X)$ (the general case will follow since if a sequence $y_k$ converges to $y$, then writing $Y = D(X) \oplus Y'$ where $Y'$ is finite-dimensional as above with the basis, we can write $y_k = y'_k + d_k$, $y = y' + d$ and $d_k \to d$, $\phi(y_k) = \phi(d_k)$, $\phi(y) = \phi(d)$). Since $\phi_n$ is continuous, $\phi_n(d_k) \to \phi_n(d)$. Pick a sequence $x_k$ with $D(x_k) = d_k$ and some $x \in X$ with $D(x) = d$. Then
$$ |\phi(d_k) - \phi(d)| = |\phi(d_k) - \psi(x)| \le |\phi(d_k) - \phi_n(d_k)| + |\phi_n(d_k) - \phi_n(d)| + |\phi_n(d) - \psi(x)|. $$ The above inequality is true for all $n$, so we can take its $\mathrm{liminf}$ when $n$ goes to infinity to get an upper bound that depends only on $k$. The first term goes to $0$ with $n$ by definition of $\phi$, the second term goes to $0$ with $k$ by continuity of $\phi_n$, and the last term is $|\phi_n(D(x)) - \psi(x)|$, which also goes to $0$ with $n$ since $D^*\phi_n \to \psi$ in $X^*$. This means $\phi$ is also continuous and therefore $\phi \in Y^*$.
($\Leftarrow$) Suppose $D^*$ has finite dimensional kernel. Consider the vector space $Y/D(X)$. The universal property of the quotient map says that $(Y/D(X))^* \simeq \ker D^*$ via the dual of the quotient map $\pi : Y \to Y/D(X)$. It follows that $(Y/D(X))^*$ has dimension $n$. Therefore, $Y/D(X)$ also has dimension $n$, and it follows that $D(X)$ is a closed subspace of $Y$. (To see the latter, take a basis $\{\omega_1,\cdots,\omega_n\}$ of $(Y/D(X))^*$ and lift those functionals to $\{\tilde{\omega}_1,\cdots,\tilde{\omega_n}\}$ in $Y^*$. An elementary argument shows that $\bigcap_{i=1}^n \ker \omega_i = D(X)$, showing that $D(X)$ is closed; hint: consider $K = \bigcap_{i=1}^n \ker \omega_i$ and show that $(K/D(X))^* = 0$.)
Let me know what you think of this argument, I think it's still very elementary but I can add some details if you like.