# A non-trivial example of a non-surjective bounded linear operator that is not open?

A bounded linear operator $A:X\to Y$, where $X$ and $Y$ are Banach spaces, is open, if it is surjective. Consider an operator $B:\mathbb{R} \to \mathbb{R}$ that is not surjective. Therefore, it may not map open sets to open sets.

But I am struggling to think of a non-trivial explicit example of $B$? If I take $B$ to be given by $x \to B(x) = \text{const}$, I have a non-open set. But is this trivial example the only type of function we can have for $B$? If I take $B$ to be any common function, it seems any open interval will still get mapped to an open interval. What is an explicit non-trivial example for $B$ that will not take an open set to an open set.

• I'm confused about what you're asking. Is $B$ supposed to be a linear operator from $\mathbb R$ to $\mathbb R$ that is not surjective? Not too many of those... – Robert Israel Jul 28 '17 at 6:58
• @RobertIsrael I was just trying to pick the simplest space, but if that is too simple, then an example is a slightly more complicated space is fine. – eurocoder Jul 28 '17 at 8:26

An example exists already for the plane, $B:\Bbb R^2\to \Bbb R^2$, $(x,y)\mapsto x$.
On the other hand, each open (even not necessarily bounded) linear operator $B$ between topological vector spaces (say, over $\Bbb R$) $X$ and $Y$ is surjective, because the image $B(X)$ is an open linear subspace of $Y$, so it coincides with $Y$.
You can show that all such functions must be in some sense like your or Alex Ravsky's functionals: surjective on some subspace, and constant on the other. Take the subspace that forms $B$'s image; call this $Z$. Then take a complement to that space (the cokernel); this can be written $W = Y/Z$. By definition $B$'s projection onto $W$ is a constant (zero), and $B$ is surjective on $Z$ (and any projections onto $Z$ are trivial, since it's already all inside $Z$). So $B$ can be described as the tensor sum of two maps: one surjective and one constant zero.
In particular these two subspaces have to add in dimension to the dimension of the parent space. That means that in the $\mathbb{R}\to\mathbb{R}$ case, indeed everything is either of the form $B(x) = ax$ with $a\neq 0$, or $B(x) = 0$ (but you knew that already!), so that your example is the only one for that specific pair of spaces.
• It is naturally consider a complement space to $Z$ as a subspace of $Y$, especially when we are talking about a representations of $B$ as a tensor sum of two maps, that is essentially about maps $B_Z:B\to Z$ and $B_W:B\to W$ such that $B(x)=B_Z(x)+B_W(x)$ for each $x\in X$, right? Then why such subspace $W$ exists for $Z$? Also $\operatorname{dom} B=X$ so $B$ may be undefined on $W$ and then the sentence “$B$ is constant on $W$ may be incorrect. – Alex Ravsky Jul 28 '17 at 7:29
• $Z$ and $W$ are both supposed to be subspaces of the image, not the domain. There are indeed maps $B_Z : B \to Z$, but this is just $B : X \to Y$ followed by $\proj_Z : Y \to Z$. The punchline is that $B_Z$ is identical to $B$ as a function, just as a map it has a smaller associated space; and that $B_W$, the projection of $B$ onto $W$, is the constant zero function $X \to W$. That's what I meant by "constant on $W$" although I realize that might be unclear -- I'll change that wording. – Alex Meiburg Jul 28 '17 at 7:34
• OK, but you don't need to use $\operatorname{proj}_Z$ because $B(X)\subset Z$ already. – Alex Ravsky Jul 28 '17 at 7:43