0
$\begingroup$

Let $\mathcal{B}$ be a Banach space, and $V$ a normed linear space. $L_0,L_1:\mathcal{B}\to V$ are bounded linear operators. Assume $\exists c$ such that $L_t := (1-t)L_0 + tL_1$ satisfies:

$$(*)\quad \|x\|_\mathcal{B} \leq c\|L_t x\|_V,\quad \forall t\in [0,1].$$

Proposition: $L_0$ is onto $\iff$ $L_1$ is.


I want to prove this. So I can see if we assume $L_s$ is surjective for some $s\in[0,1]$ we get that $L_s$ is bijective from linearity and $(*)$ and we get an inverse operator from $(*)$ and surjectivity, $L_s^{-1}$.

I don't know how to proceed though. Can I have some guidance please.

$\endgroup$
  • $\begingroup$ If there are site related quirks I should know about to make my question more 'findable' and/or fit better with the standards, please let me know in this comments box. $\endgroup$ – Jack Don Oct 24 '15 at 15:48
  • $\begingroup$ what is x in your inequality (*) ? $\endgroup$ – DanielWainfleet Oct 24 '15 at 16:33
  • $\begingroup$ @user254665 it was missing on the rhs. $\endgroup$ – Thomas Oct 24 '15 at 16:42
1
$\begingroup$

The point of the standard proof to this (which can be found, e.g., in Gilbarg Trudingers 'Elliptic Partial differential equations of Second order', Theorm 5.2) is to assume that $L_s$ is onto for some $s$ and then show that this implies $L_t$ is for all $t$ in a controlled neighbourhood of $s$. By covering $[0,1]$ with finitely many such intervals you can show that the set where $L_s$ is onto is connected.

To show the statement I indicated above you can reformulate the equation $L_t x = y$ as a fixed point problem for $T$ given by $$Tx := L_s^{-1}y + (t-s) L_s^{-1}(L_0 - L_1) x $$

Now for that $T$ you can verify the prerequisites of the Banach fixed point theorem if $t$ is close enough to $s$.

If this is not enough hint for you you can always look it up (see the source mentioned above)

(Edit: as indicated, the point is always to show that the set where some property for some continuos family of operations you want to be true is connected. This usually means you want to show it's open and closed. Often, open is more or less easy, since, for example, the set of invertible maps is often open in reasonable setups. To show that the set is closed, you usually try to apply some compactness result. This is where, e.g., in elliptic PDE the compact embeddings of Sobolev spaces or the Schauder estimates kick in. In the proof I referred to applying compactness is somewhat hidden behind the clever observation that the applicability of the Banach fixed point theorem can be shown to be true on intervals of controlled size (controlled by the norms of $L_0$ and $L_1$, to be more precise)).

$\endgroup$
  • $\begingroup$ I will check out the proof, thank you for the reference. Would it be possible for me to ask you additional queries about it once I do? Just in the case that I don't understand something I mean. $\endgroup$ – Jack Don Oct 24 '15 at 16:25
  • $\begingroup$ @JackDon If you end up with a new question you should consider creating a new question on this site. $\endgroup$ – Thomas Oct 24 '15 at 16:34
  • $\begingroup$ @JackDon also, you asked for a hint. I tried to formulate my answer accordingly. If you were serious about that just try to see why $Tx = x$ is equivalent to $Lx = y$, with given $y$, and then try to show that $T$ is a contraction. It's not too difficult. This assumes you now the Banach fixed point theorem, of course. $\endgroup$ – Thomas Oct 24 '15 at 16:40
  • $\begingroup$ Oh yes this is greatly appreciated thanks. I didn't mean for further queries to mean that I didn't appreciate what you have written. $\endgroup$ – Jack Don Oct 24 '15 at 16:57
  • $\begingroup$ You mention connectedness twice. You are referring to the topological condition of being a connected space, as opposed to being a seperable space? Also I can't see connectedness or compactness come into play in the recommended proof, through Banach fixed point theorem or otherwise. Mainly the step where we choose that $T:\mathscr{B}\to\mathscr{B}$ map I am not sure what we are doing $\endgroup$ – Jack Don Oct 25 '15 at 5:30

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.