I watched videos on the Cauchy-Schwarz inequality, but don't know how to work this problem due to the $\mathbf{a}^\top \mathbf{x}$ part.
Let $\mathbf{a}\in \Bbb R^n$ be given. Define $f: \Bbb R^n\to\Bbb R$ by $f(\mathbf{x}) = \mathbf{a}^\top \mathbf{x}$. Show that $f$ is continuous. Hint: Use the Cauchy-Schwarz inequality
If someone can show steps/explain that would be great. Thank you!
The dot product of two vectors $u,v \in \Bbb R^n$ may be written in matrix notation as $u^\top v$, where $u$ and $v$ are now seen as column vectors. Then Cauchy-Schwarz reads $|u^\top v| \leq |u||v|$. That said, if $a = 0$ then $f = 0$ is continuous. If $f \neq 0$, let $\epsilon > 0$ and take $\delta = \epsilon/|a|>0$. If $|x-y|<\delta$, then we have that $$|f(x)-f(y)| = |a^\top x - a^\top y| = |a^\top(x-y)| \stackrel{(\ast)}{\leq} |a||x-y| < |a|\delta = \epsilon,$$where in $(\ast)$ we have used Cauchy-Schwarz. So $f$ is, in fact, uniformly continuous.