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$X$ is a normed vector space and $T \in X^*$. Show that \begin{equation} \label{prob} T \in X' \Leftrightarrow N(T) \; \text{is closed}, \end{equation} where $N(f)$ is the null space of $T$.

i) Prove that if $T \in X'$, then $N(T)$ is closed.

ii) Assume that $T$ is unbounded. Show that then $N(T)$ is not closed.

iii) Make an argument that (i) and (ii) together prove \eqref{prob}.

My current suggestions.

i) Let $\{v_n\}_{n=1}^\infty \in N(T)$ such that $v_n \rightarrow v$. Because $v_n \in N(T)$ then $Tv_n = 0$. Now $T \in X'$ is bounded and linear giving $\lim_{n \rightarrow \infty} Tv_n = Tv$ giving us $Tx = 0$. Therefore we must have $v \in N(T)$. Thus $N(T)$ is closed (I got this conclusion from a theorem in a book by Kreyszig).

ii) I'm not sure here, but my idea is to take a sequence $v_n \in X$ converging to $v$ and a sequence $y_n \in N(T)$, and then prove that $y \rightarrow v$ thus proving that $N(T)$ is not closed. I don't know exactly how this should be done.

iii) This I don't know but since I haven't used that $T \in X^*$ I have a feeling that I should use this?

To clarify: $X^*$ is the normed space of all linear functionals and $X'$ is the normed space of all linear and bounded functionals.

ANYN11
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  • To clarify, is $X^*$ the space of all linear maps $X \to \mathbb{C}$ (or whatever the base field is), and $X'$ the subspace of bounded/continuous ones? – arkeet Sep 26 '16 at 16:35
  • I see how that could be unclear. I will clarify in the post. Thanks – ANYN11 Sep 26 '16 at 16:37
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    Isn't it enough to say that ${0}$ is closed and $T$ is continuous so $T^{-1}({0}) = \ker T$ is closed? For part i), that is. – Eman Yalpsid Sep 26 '16 at 16:43
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    @Andrew Sure it is. The proof used in the question is basically the proof of the equivalence between the two notions of continuity ("topological" and sequential) for metric spaces. – Daniel Robert-Nicoud Sep 26 '16 at 16:46

1 Answers1

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You can prove (ii) by contradiction: Suppose $N(T)$ is closed, then $Y:= X/N(T)$ is a normed linear space, and the quotient map $$ \pi : X\to Y $$ is a bounded linear map. Now, by the first isomorphism theorem, $T$ induces an injective map $$ \overline{T} : Y\to \mathbb{C} $$ Furthermore, $Y$ is finite dimensional (in fact, one dimensional) because if $y_0 \notin N(T)$, then it is an easy check that $$ Y = \text{span}(y_0 + N(T)) $$ Since any two norms on $Y$ are equivalent, it follows that $\overline{T}$ is also continuous. Hence, $$ T = \overline{T}\circ \pi $$ must also be continuous. This contradiction proves (ii).

Prahlad Vaidyanathan
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