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 $V$?

  • $\begingroup$ 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

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 $H=P(H)+V$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.