# Compact operators, injectivity and closed range

Let $X$ be a an infinite dimensional Banach space. $A\in B(X)$ is a compact operator. If its range $\operatorname{Im}(A)$ is closed in $X$ then $A$ cannot be injective because $A:X\to \operatorname{Im}(A)$ would be a compact bijection between Banach spaces and the unit ball $B_X=A^{-1}AB_X$ would be compact.

Now if $A$ is not injective, can we say that $\operatorname{Im}(A)$ must be closed ? Or if this is false, can we find a non injective compact operator with non closed range (i.e. infinite dimensional range) ?

• Note that "closed" and "finite dimensional" are not the same thing. There exist infinite dimensional subspaces that are closed, as well as infinite dimensional subspaces that are not. – Giuseppe Negro May 23 '14 at 22:06
• @GiuseppeNegro But I think "closed" and "finite dimensional" range are the same thing for a compact operator in Banach spaces. – user165633 May 23 '14 at 22:23
• You are absolutely right. I also updated the example in my answer, now it should match your question better. – Giuseppe Negro May 24 '14 at 8:13

Consider the operator $T\colon \ell^2\to \ell^2$ defined by $$T\mathbf{x}=\left(0, \frac{x_2}{2}, \frac{x_3}{3}\ldots\right).$$ $T$ is not injective because $T\mathbf{e}_1=0$, but it is compact and its range is the subspace $$E=\left\{\mathbf{y}\in \ell^2\ :\ y_1=0,\ (ny_n)\in \ell^2.\right\}.$$ The subspace $E$ is dense in $M=\overline{\mathrm{span}}\{\mathbf{e}_2, \mathbf{e}_3\ldots\}$, because it contains $\{\mathbf{e}_2, \mathbf{e}_3, \ldots\}$, but the inclusion $E\subset M$ is proper. For example, the sequence $$\mathbf{x}=\begin{cases} 0, & n=1 \\ \frac{1}{n}, & n \ge 2 \end{cases}$$ does not belong to $E$. Therefore $E$ is not closed in $M$, hence it is not closed in $\ell^2$ too.

P.S. In comments the OP requested an example in $C([0,1])$ space. This of course is more complicated because we have now left the elementary Hilbert space setting. Indeed, I am not able to furnish an example, nor I think that a natural example exists. Let me explain what natural means to me in this context.

The typical examples of compact operators in $C([0,1])$ space are integral transforms $$Tf=\int_0^1 K(x, y) f(y)\, dy.$$ In turn, integral transforms usually arise as solution operators to linear boundary value problems for ODEs: $$\begin{cases} Lu = f, & x\in (0, 1) \\ U(u)=0\end{cases}$$ Here $L$ is a symmetric linear differential operator and $U(u)=0$ denotes the boundary conditions. (Notation is taken from Coddington-Levinson's book on ordinary differential equations, chapter 7). Now such an operator is always injective, as $Tf=0$ means that the solution to the boundary value problem is the null function $u=0$, which forces $f=Lu=L0=0$. So an example cannot come from there.

There are also integral transforms $T$ that do not come from a boundary value problem, of course. However, the typical examples here have splitting kernel: $$K(x, y)=\sum_{j=1}^J a_j(x)b_j(y).$$ This splitting leads to operators with finite dimensional range. So an example cannot come from there either.

That's what I meant with natural above. But there are a lot more compact operators on $C([0, 1])$. I am sure that an example of an operator which is compact, not injective, and not finite rank can be easily crafted, perhaps with tools of abstract functional analysis (such as Schauder bases).

• What I want is a compact non injective operator with non closed range. – user165633 May 23 '14 at 22:26
• Ok. A small modification of this example will do. – Giuseppe Negro May 24 '14 at 8:04
• @user144542: Now it should be OK. – Giuseppe Negro May 24 '14 at 8:11
• That answers my question, thank you. – user165633 May 24 '14 at 12:06
• Can we find an example in the space $X=C[0,1]$ of continuous real function on $[0,1]$ ? Every time I find a compact operator on $X$ which is not injective, I see its range and I find it closed (finite dimensional), that's the reason behind my question here. Or maybe this deserves to be in a new question. – user165633 May 24 '14 at 16:12

A compact linear operator $A\colon X \to Y$, where $X$ and $Y$ are Banach spaces, has closed range if and only if it has finite rank.

Because if the range is closed, it is a Banach space, and the open mapping theorem asserts that $A(B_X)$ is a neighbourhood of $0$. But its closure is compact, so $\mathcal{R}(A)$ is locally compact. By Riesz' theorem, a Hausdorff topological vector space is locally compact if and only if it is finite-dimensional.

• I know this, this is actually what I said in the last line of the original post. Maybe my question is miss-understood. – user165633 May 23 '14 at 22:29
• The point is that injectivity has absolutely nothing to do with it. You ask "Now if $A$ is not injective, can we say that $Im(A)$ must be closed ?" And I say: "No, the range is closed if and only if it is finite-dimensional, regardless of whether the operator is injective". – Daniel Fischer May 23 '14 at 22:37
• I thought there was a link to injectivity because: a compact operator $A$ with closed range cannot be injective. – user165633 May 23 '14 at 22:49
• Well, unless the domain is finite-dimensional. A compact operator has closed range if and only if its kernel has finite codimension. – Daniel Fischer May 23 '14 at 22:54
• Thank you for the information. I am now satisfied with the counterexample given by Giuseppe Negro. – user165633 May 24 '14 at 12:16