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Let $T:X\rightarrow Y$ be a linear operator such that $Y$ is finite dimensional. Show that $T$ is continuous if and only if Kernel of $T$ is closed in $X$.

Suppose $T$ is discontinuous. There exists $x_n\in X$ such that $||x_n||=1$ such that $||Tx_n||$ diverges.

As $T$ is a nonzero operator, there exists $a\in X$ that is not in kernel of $T$.

We look for elements of kernel of $T$ that converges to $a$.

Consider the case when $Y$ is of dimension $1$. Let $a_n=a-\frac{T(a)x_n}{T(x_n)}$. See that $a_n$ is in kernel of $T$ and $a_n\rightarrow a$ as $x_n$ is of norm $1$ and $Tx_n$ diverges.

So, we have $a_n$ in kernel of $T$ and $a_n$ converges to $a$ but $a$ is not in kernel which concludes that Kernel is not closed.

I am not able to extend this to case of arbitrary finite dimension $Y$

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Suppose $T : X \to Y$ is continuous. Then $\text{ker}(T) = T^{-1} (\{0\})$ is closed.

Conversely, suppose $\text{ker}(T)$ is closed in $X$. Then the quotient space $X / \text{ker}(T)$ is a normed space. Define the map $S : X / \text{ker}(T) \to Y$ by $x + \text{ker}(T) \mapsto T(x)$. Then $S$ is continuous since it is an operator on a finite-dimensional space. Note that $T = S \circ \pi$, where $\pi : X \to X / \text{ker}(T)$ is the quotient map. Thus $T$ is bounded as the composition of bounded operators.

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  • $\begingroup$ This is brilliant.... I don't find any gaps as of now.. $\endgroup$ – user312648 Aug 15 '16 at 14:37
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If you don't want to use quotients, you can argue as follows.

I assume that the scalar field is $\mathbb C$. Note that it suffices to show that $T$ is bounded in some neighborhood of $0$.

If $\ker T$ is closed and not all of $X$, there is an $x\in X$ and a balanced neighborhood $U$ of $0$ s.t.

$\tag1 (x+U)\cap \ker T=\emptyset.$

Now, $TU$ is balanced because $T$ is linear, so either $TU$ is bounded or $TU=\mathbb C$. In the latter case, we can find an $x_1\in U$ s.t. $-Tx=Tx_1$, which says that $x+x_1\in \ker T$, a contradiction. We conclude that $T$ is bounded, and the result follows.

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