Problem of the Week #107 - June 16th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: For normed linear spaces $X$ and $Y$, let $T:X\rightarrow Y$ 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.

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[Chris L T521]

This week's problem was correctly answered by Euge. You can find his solution below.

[sp]Suppose $X$ is finite dimensional. Let {$e_1, e_2, \ldots, e_m$} be a basis for $X$. Given $x \in X$, there are unique scalars $x^1, x^2,\ldots, x^m$ such that $x = \sum_{i = 1}^m x^i e_i$. By linearity of $T$ and the triangle inequality,

$\|Tx\|_Y = \left\|\sum_{i = 1}^m x^i Te_i\right\|_Y \le \sum_{i = 1}^m |x^i| \|Te_i\|_Y \le c \|x\|_1$,

where $c = \max\{\|Te_1\|_Y, \|Te_2\|_Y,\ldots, \|Te_m\|_Y\}$ and $\| \cdot \|_1$ is the 1-norm with respect to the given basis for $X$. Since $X$ is finite dimensional, $\| \cdot \|_1$ is equivalent to $\| \cdot \|_X$. So $\|x\|_1 \le d\|x\|_X$ for some constant $d$ independent of $x$. Hence $\|Tx\|_Y \le cd \|x\|_X$. Since $x$ was arbitrary and constants $c$ and $d$ are independent of $x$, it follows that $T$ is continuous.

Next, suppose $Y$ is finite dimensional. Assume $T$ is continuous. For a given $x$ in the closure of ker $T$, there exists a sequence $(x_n)$ in $X$ such that $\lim_n x_n = x$. Since $Tx_n = 0$ for all $n$, continuity of $T$ yields $Tx = \lim_n Tx_n = 0$, i.e., $x\in \text{ker} T$. Thus ker $T$ is closed.

Conversely, suppose ker $T$ is closed. Then the quotient space $X/\text{ker}\, T$ is a normed linear space with norm $\|x + \text{ker}\, T\| := \inf\{\|x - u\|: u \in \text{ker} T\}$. As a vector space, $X/\text{ker}\, T$ is isomorphic to a subspace of the finite dimensional vector space $Y$, so $X/\text{ker}\, T$ is finite dimensional. Furthermore, linearity of $T$ implies linearity of the induced map $X/\text{ker}\, T \to Y$. So by the argument given in the first paragraph, the induced map is continuous. The natural projection map $\pi : X \to X/\text{ker}\, T$ is continuous since $\|\pi(x)\| \le \|x\|_X$ for all $x \in X$. Since the composition of continuous maps is continuous and $T$ is the composition $X \to X/\text{ker}\, T \to Y$, it follows that $T$ is continuous.[/sp]