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I am reading a paper on PDEs and at some point the author uses an argument that I cannot understand very clear, it seems to be elementary but I do not get it.

Let $P$ be a Fredholm operator (i.e. finite dimensional kernel and cokernel) acting between Hilbert spaces $P:H \longrightarrow F$. Suppose that $\mathfrak{A}$ is a dense linear subspace (not closed) of $F$. For example $\mathfrak{A}=\mathcal{C}^{\infty}_{0}(\mathbb{R}^n)$ and $F=L^{2}(\mathbb{R}^n).$ Then the argument says that there exits a finite dimensional space $V\subset \mathfrak{A}$ such that $F=P(H)+V$.

On one side we know that $F=P(H)\oplus \text{Coker P}$ as a direct sum and because $P$ is Fredholm we have that $\text{Coker P}$ is finite dimensional, but in general $\text{Coker P}$ is NOT a subset of $\mathfrak{A}$.

The argument says that somehow we can choose $V\subset \mathfrak{A}$ such that we trade $\text{Coker P}$ by $V$ and we obtain a decomposition $F=P(H)+V$ (that is no longer direct sum in general?).

How can we choose $V$?

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  • $\begingroup$ As stated, the assertion is wrong. The dense set $\mathfrak A$ may not contain any subspace at all (for instance, think $\mathfrak A=F\setminus\{0\}$. $\endgroup$ – Martin Argerami Apr 25 '16 at 22:04
  • $\begingroup$ You are right, actually $\mathfrak{A}$ is also a linear space, I forgot to mention that. So $\mathfrak{A}$ is a dense linear space in $F$. For example $\mathcal{C}^{\infty}_{0}(\mathbb{R}^n)$ in $L^{2}(\mathbb{R}^n)$. $\endgroup$ – Coffee Apr 25 '16 at 22:13
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Let $e_1,\ldots,e_k$ be an orthonormal basis of $\text{Coker}\,P$.

Note that $P(H)$ is closed, so there exists $f_1\in\mathfrak A$ with $$ \text{dist}\,(f_1,e_1)<\frac{\text{dist}\,(e_1,P(H))}2. $$ The triangle inequality guarantees that $f_1\not\in P(H)$. Now $\text{span}\,\{P(H),f_1\}$ is closed, and we can repeat the process to obtain $f_1,\ldots,f_k$, as long as $\text{span}\,\{P(H),f_1,\ldots,f_{k-1}\}$ is a proper subspace of $H$.

Let $V=\text{span}\{f_1,\ldots,f_k\}\subset\mathfrak A$. If $k<n$, this means that the process stopped, so $P(H)+V=H$. If $k=n$, we can use this result to show that $\dim H/(P(H)+V)=0$, so $H=P(H)+V$.

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  • $\begingroup$ Thanks Martin, good argument. There is only one point I am not sure. If $k=n$ in order to use the result you linked we need to have $V\cap P(H)=\varnothing $ i.e. we need to have a direct sum $P(H)\oplus V$. How can we guarantee that the span of $\lbrace f_1 ,...f_n \rbrace$ will not intersect $P(H)$. My guess is that even though we do not have direct sum, it is possible to show that $H=P(H)+V$ but I do not see the argument, what do you think? $\endgroup$ – Coffee Apr 26 '16 at 20:51
  • $\begingroup$ By construction, $f_1,\ldots,f_n$ are linearly independent with $P(H)$ (that's the whole point of how I constructed the $f_j$). $\endgroup$ – Martin Argerami Apr 26 '16 at 20:53
  • $\begingroup$ That is right, I had to think a little bit harder to see the linear independence, but now it is clear. Thanks for your answer Martin. $\endgroup$ – Coffee Apr 26 '16 at 22:26
  • $\begingroup$ Glad I could help. $\endgroup$ – Martin Argerami Apr 26 '16 at 23:07

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