Find a positive constant $C_1 \geq 0$ such that ....

Question

Here $x = (x_1,x_2) \in \mathbb{R}^2$,

Consider the matrix $$A= \ \left( \begin{array}{ccc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right)$$ as a linear ma from $\mathbb{R}^2$ to $\mathbb{R}^2$.

Find a positive constant $C_1 \geq 0$ such that $$\|Ax\|_1 \leq C_1\|x\|_1$$ for all $x \in \mathbb{R}^2$. Conclude From here that the linear map $A$ is continuous.

Now I know that you must use $$\|Ax\|_1 = |a_{11}x_1+a_{12}x_2|+|a_{21}x_1+a_{22}x_2|$$ to get to $$=(a_{11}+a_{21})|x_1|+(a_{12}+a_{22})|x_2|$$ $$\leq M(|x_1|+|x_2|) = M\|x\|_1$$

From this point on I am unsure how to determine $C_1=M$.

Thanks in advance.

Answer 1

$\|Ax\|_1 = |a_{11}x_1+a_{12}x_2|+|a_{21}x_1+a_{22}x_2| \le |a_{11}||x_1|+|a_{12}||x_2|+|a_{21}||x_1|+|a_{22}||x_2|$

Since $|x_1|,|x_2| \le ||x||_1$, we get, with $C_1:= \max\{|a_{11}|,|a_{12}|,|a_{21}|,|a_{2}|\}$:

$\|Ax\|_1 \le ||x||_1$