Bounded linear functional considering a $1 \times 2$ matrix

Question

Consider the $1\times2$ matrix $A = [2\; -3]$.

Show that the mapping $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $f(\vec{x}) = A\vec{x}$ is a bounded linear functional using 2-norm.

Attempt:

Let $\alpha \in \mathbb{F}$ such that

$f(\alpha\vec{x}) = A (\alpha\vec{x})$

$=[2 \quad -3] [\alpha x_{1} \quad \alpha x_{2}]^T$

$=[2\alpha x_{1} \quad -3\alpha x_{2}]$

$= \alpha [2 x_{1} \quad -3 x_{2}]$

$=\alpha [2 \quad -3] [x_{1} \quad x_{2}]^T$

$=\alpha A \vec{x}$

$=\alpha f(x)$

Now, consider,

$f(\vec{x} + \vec{y}) = A(\vec{x}+\vec{y})$

$= [2 \quad -3]([x_{1} \quad x_{2}]^T + [y_{1} \quad y_{2}]^T)$

$=[2 \quad -3]([x_{1} + y_{1} \quad x_{2} + y_{2}]^T)$

$=2x_{1} + 2y_{1} \quad -3x_{2} + -3y_{2}$

$=2x_{1} - 3x_{2} \quad 2y_{1} + -3y_{2}$

$=[2 \quad -3] [x_{1} \quad x_{2}]^T + [2 \quad -3] [y_{1} \quad y_{2}]^T$

$=A\vec{x} + A\vec{y}$

$=f(x) + f(y)$

Hence, the mapping $f$ satisfies the linear condition. The second condition required in order to prove that the mapping is a bounded linear functional is as follows:

$2)$ There exists a constant $c_{f}$ such that

$|f(\vec{x})| \leq c_{f}\|\vec{x}\|$

$\vec{x} \in \mathcal{X}$, where $\mathcal{X}$ is a normed linear space.

I don't really know where to start with for this condition. Any help would be greatly appreciated.

Answer 1

$f(x_1,x_2)=2x_1-3x_2$. Your have proved that $f(\alpha (x_1,x_2))=\alpha f((x_1,x_2))$ but you also have to prove that $f((x_1,x_2)+(y_1,y_2))=f(x_1,x_2)+f(y_1,y_2)$. Can you do that?

Answer 2

For boundedness, note that $$ |f(x,y)|=|2x-3y|\leq 2|x|+3|y|\leq 3(|x|+|y|)\leq 3\sqrt{2}\sqrt{x^2+y^2} $$

Where the final inequality follows from $$ 0 \leq (|x|-|y|)^2=x^2+y^2-2|x||y|\implies 2|x||y|+x^2+y^2\leq2(x^2+y^2)\\ \implies (|x|+|y|)^2\leq 2(x^2+y^2)\\ \implies |x|+|y|\leq\sqrt{2}\sqrt{x^2+y^2} $$

Alternatively, use Cauchy Schwarz, $$ |(2,-3)\cdot (x,y)|\leq ||(2,-3)||\cdot||(x,y)||=\sqrt{13}||(x,y)|| $$ This maximum is achieved by taking $(x,y)$ parallel to $(2,-3)$. Note that $\sqrt{13}<3\sqrt{2}$ and our first strategy was far from optimal.