# a problem using the Cauchy-Schwarz inequality

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.