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Let $\vec v\in\mathbb R^2$. Prove that the linear map $L\colon\mathbb R^2\to \mathbb R$ given by $L(\vec x)=\vec v^T\vec x$ is continuous.

I have a question about my proof:

Consider $\vec x_0\in\mathbb R^2. $Let $\epsilon>0$ and let $\delta=\frac{1}{\lVert\vec v^T\rVert}$. Then $$\lVert \vec v^T\vec x-\vec v^T\vec x_0\rVert =\lVert \vec v^T(\vec x-\vec x_0)\rVert\leq\lVert \vec v^T\rVert \lVert \vec x-\vec x_0\rVert<\epsilon. $$ Am I allowed to use the norm of $\vec v^T$? Is it true that $$ \vec v\cdot \vec v=\vec v^T\cdot \vec v^T, $$ where we use the standard dot product?

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    What even *is* $v^T\cdot v^T$? Surely you aren't implying that the dot product on the left of the equation is the *same* dot product on the right... – 2017-02-19
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    $| a^T\cdot b|≤\|a\|\,\|b\|$ is also known as the Cauchy-Schwarz inequality. – 2017-02-19

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