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I've read somewhere that the kernel of a linear map is closed iff the map is bounded.

Consider the derivative operator $D: \mathcal C^1([0,1],\mathbb C) \to \mathcal C([0,1],\mathbb C)$, i.e. $Df = f'$.

Since this map is not bounded, the kernel must not be closed.

I'm a bit confused. I thought the kernel of this map was the constant functions. I don't see why this kernel wouldn't be closed.

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    $\begingroup$ With suitable Banach topologies, the derivative map is certainly continuous (a.k.a. "bounded"), and the kernel, constant functions, is certainly closed. The "correct" norm on once-continuously differentiable functions is sup norm of the absolute value of the function + sup norm of absolute value of the derivative. So it is immediate that "derivative" is continuous. $\endgroup$ – paul garrett Nov 17 '12 at 20:20
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    $\begingroup$ The kernel of a linear map is closed iif the map is bounded : Suppose that $T$ is a continuous linear map. There exist $x_n \subseteq X$ with $\|x_n\| = 1$ for all $n$ such that $\|T(x_n)\| \rightarrow \infty$. If $a \not\in$ ker$(T)$. Let $x'_n := a - \frac{T(a)}{T(x_n)}x_n$. Then $T(x'_n) = T(a) - \frac{T(a)}{T(x_n)}T(x_n) = 0$. So $x'_n \in$ ker$(T)$ and $x'_n \rightarrow a$. $\endgroup$ – Joe G. Nov 17 '12 at 20:49
  • $\begingroup$ I shouldn't have deleted my comment. You argument works only for range $\mathbb R$ ($\mathbb C$ respectively), but not in general (you are dividing by $T(x_n)$, what is this for general range space?). The implication: "Kernel closed $\Rightarrow T$ continuous" holds true in general only for finite dimensional range space. (The constants are closed, since they are finite dimensional, hence complete). $\endgroup$ – martini Nov 17 '12 at 20:59
  • $\begingroup$ I would be interested in seeing where you read that a closed kernel implies continuity of the operator. I'm certain that's the case for finite-rank operators, but where have you seen the general assertion? $\endgroup$ – Martin Argerami Nov 17 '12 at 21:30
  • $\begingroup$ Martini, I know that you must be right, but I'm still not sure to understand. $\endgroup$ – Joe G. Nov 17 '12 at 21:44
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Your assumption is incorrect. It is not true that a closed kernel implies continuity of the operator. It is indeed true for finite-rank operators, but not in general.

As an example -- beyond yours -- let $X=P[0,1]$, i.e. polynomials on $[0,1]$ with the max norm (i.e. $\|\cdot\|_\infty$). Let $T$ be the operator $p\mapsto Tp$, where $Tp(t)=p(t+1)$. It is clear that $T:X\to X$ is a linear bijection, so $\ker T=\{0\}$. But $T$ is not continuous: let $p_n$ be a sequence of polynomials that converge to the continuous function $$ f(t)=\begin{cases}0,&\text{ if }t\leq1\\ t-1,&\text{ if }t\geq1\end{cases} $$ Then $p_n\to0$ in $X$, while $\|Tp_n\|=1$ for all $n$.

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  • $\begingroup$ Is there an counterexample $T:X\to Y$ with $X,Y$ Banach spaces and $T$ defined everywhere? $\endgroup$ – Jose27 Nov 17 '12 at 22:04
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    $\begingroup$ @Jose27: Take countable linearly independent set of unit vectors $e_n$ in $X$. Let $T e_n = n e_n$ and let $T$ be the identity on an algebraic complement of the span of the $e_n$. Then $T$ is injective, defined everywhere and unbounded. There's no way to avoid the axiom of choice, hence an example is necessarily unnatural and non-explicit, see here and here. $\endgroup$ – commenter Nov 17 '12 at 22:21

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