TITLE: $T$ is continuous if and only if $\ker T$ is closed
QUESTION [36 upvotes]: Let $X,Y$ be normed linear spaces. Let $T: X\to Y$ be linear. If $X$ is finite dimensional, show
that $T$ is continuous. If $Y$ is finite dimensional, show that $T$ is continuous if and only if $\ker T$ is closed.    
I am able to show that $X$, finite dimensional $\implies$ $T$ is bounded, hence continuous.   
For the second part: This is what I have:  
Suppose $T$ is continuous. By definition $\ker T = \{ x\in X : Tx = 0 \}$ , and so $\ker T$ is the continuous inverse of a closed set. Hence $\ker T $ is closed.  
First, is what I have attempted okay. How about the other direction?

REPLY [15 votes]: I am doing for $Y=\mathbb{R}$; 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.