TITLE: Showing that $\ker T$ is closed if and only if $T$ is continuous.
QUESTION [27 upvotes]: Possible Duplicate:
$T$ is continuous if and only if $\ker T$ is closed 


Let $T: X\to \mathbf{R}$ be linear. Suppose that $X$ is a Banach space. I want to show that $T$ is continuous if and only if $\ker T $ is closed.  

My Attempt. 
$(\Rightarrow)$ Suppose $T$ is continuous. Then if $x_n\to x$, then $T(x_n)\to T(x)$. Let $x_n \in \ker T$. Then $T(x_n) = 0$. Using continuity, $$ T(x) = \lim_{n\to \infty} T(x_n) = 0.$$ Hence $x\in \ker T$ and thus $\ker T$ is closed.  
$(\Leftarrow)$  Suppose $T$ is not continuous. So $T$ is not bounded. i.e. $\exists$ a sequence $x_n$ such that $T(x_n) \to \infty$ as $n\to \infty$. Let $a\notin \ker T$. Then defining $$x_n' = a - \frac{T(a)}{T(x_n)}x_n ,$$ it is clear that $T(x_n') = 0$ and so $x_n'\in \ker T$. Also $x_n' \to a \notin \ker T.$ So $\ker T$ is not closed.  Hence $\ker T$ closed implies that $T$ is continuous.   
Have I approached this question correctly? Are there other ways of approaching it?

REPLY [10 votes]: Clearly if $f$ is continuous then its kernel is closed set. for the converse, assume that $f\neq0$ and that $f^{-1}(\{0\})$ is a closed set. Pick some $e$ in $X$ with $f(e)=1$. Suppose by way of contradiction $||f||=\infty$. Then there exists a sequence $\{x_n\}$ in $X$ with $||x_n||=1$ and $f(x_n)\ge n$ for all $n$. Note that the sequence $\{y_n\}$ defined by $y_n=e-\frac{x_n}{f(x_n)}$, satisfies $y_n\in f^{-1}(\{0\})$ for all $n$ and $y_n\rightarrow e$. Since the set $f^{-1}(\{0\})$ is closed it follows that $e$ must belong to it and consequently $f(e)=0$ which is a contradiction. Thus $f$ is a continuous linear functional.