3
$\begingroup$

Consider normed spaces $X$ and $Y$. You can assume that they are Banach spaces if needed. Let $\mathcal{L}(X, Y)$ denote the spaces of bounded linear operators from $X$ to $Y.$ Now consider the set

$$\Omega=\{T \in L(X,Y): T \textrm{ is onto}\}.$$ Is $\Omega$ open with the norm topology?

$\endgroup$
  • 1
    $\begingroup$ Just an idea. I'd try to take any $T\in\Omega$. Then $\|T\|>0$. Let $\epsilon=\frac12\|T\|$ and try to show that $\|T-T'\|<\epsilon$ implies $T'\in\Omega$. $\endgroup$ – ajotatxe Dec 29 '17 at 20:15
3
$\begingroup$

We've seen in another answer that this is false if $X$ is not a Banach space.

Lemma Suppose $X$ is a Banach space, $T:X\to Y$ is linear and bounded, $0<\epsilon<1$, $\delta>0$, and for every $y\in Y$ there exists $x\in X$ with $||x||\le\delta ||y||$ and $||Tx-y||\le\epsilon||y||$. Then $T$ is surjective.

Proof. Let $y\in Y$. Choose $x_0\in X$ with $||x_0||\le\delta||y||$ and $||Tx_o-y||\le\epsilon||y||$. Having chosen $x_0,\dots x_n$, let $s_n=x_0+\dots+x_n$ and choose $x_{n+1}\in X$ with $||x_{n+1}||\le\delta||y-Ts_n||$ and $||Tx_{n+1}-(y-Ts_n)||\le\epsilon||y-Ts_n||$.

Then $||y-Ts_n||\to0$, in fact $\sum||y-Ts_n||<\infty$, since $||y-Ts_{n+1}||\le\epsilon||y-Ts_n||$. And so $||x_{n+1}||\le\delta||y-Ts_{n}||$ implies that $\sum||x_n||<\infty$, so $s_n\to s$. It follows that $Ts=y$, since $T$ is continuous. QED.

Now suppose $T:X\to Y$ is surjective. The Open Mapping Theorem says that there exists $\delta>0$ such that for every $y\in Y$ there exists $x\in X$ with $Tx=y$ and $||x||\le\delta||y||$. Hence for any $T'$ we have $$||T'x-y||\le||T'-T||\,||x||\le\delta||T'-T||\,||y||.$$So if $\delta||T'-T||<1$ the lemma shows that $T'$ is surjective.

$\endgroup$
  • $\begingroup$ That was amazing! Thanks again @David C. Ulrich for your solution. How did you came up with the Lemma? Is it well known or you created it on the spot? $\endgroup$ – John D Dec 30 '17 at 10:49
  • $\begingroup$ I don't know whether it's well known. I just made it up, starting with a general principle that "close should be close enough" and thinking about how close to surjective a map would have to be to allow me to show it was surjective. The argument is certainly a standard sort of argument: Get something close to what you want, then get something that comes close to cancelling the error in the first approximation... $\endgroup$ – David C. Ullrich Dec 30 '17 at 14:11
3
$\begingroup$

$\Omega$ is not open in general, at least if $X$ and $Y$ are not assumed to be Banach.

Consider $c_{00}$, the space of all finitely-supported sequences, equipped with $\|\cdot\|_2$.

Define $A : c_{00} \to c_{00}$ as

$$A(x_1, x_2, x_3, \ldots) = \left(x_1, \frac12x_2, \frac13x_3\ldots\right)$$

$A$ is bounded and onto.

However, take any $\varepsilon > 0$ and pick $n_0 \in \mathbb{N}$ such that $\frac{1}{n_0} < \varepsilon$.

Define $A_{n_0} : c_{00} \to c_{00}$ as:

$$A_{n_0}(x_1, x_2, x_3, \ldots) = \left(x_1, \frac12x_2, \frac13x_3, \ldots, \frac1{n_0 - 1}x_{n_0 - 1}, 0, \frac1{n_0+1}x_{n_0+1}, \ldots\right)$$

$A_{n_0}$ is bounded and not onto. Futhermore, we have:

$$(A - A_{n_0})(x_1, x_2, \ldots ) = \left(0, \ldots, 0, \frac{1}{n_0} x_{n_0}, 0, \ldots\right)$$

Therefore $\|A - A_{n_0}\| = \frac{1}{n_0} < \varepsilon$.

$\endgroup$

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.