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?

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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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


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