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Let f,g be continuous maps of *S* in $\mathbb{R}^n$, to $\mathbb{R}^m$. Show the inner product h(x) = $\langle f(x), g(x)\rangle$ is continuous.

My attempt:

Since f,g are continuous, their product is a continuous mapping, so for $\epsilon \gt 0\: t.e\: \delta \gt 0$ such that $\parallel x – a\parallel \lt \delta $ implies $\parallel \langle f(x),g(x)\rangle – \langle f(a)g(a)\rangle \parallel \le\parallel f(x)g(x) – f(a)g(a)\parallel \lt \epsilon$ by schwartz inequality.

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(Adjust $n$, $m$ as appropriate, the question is a little ambiguous.)

The map $\phi: \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$ (or complex, if it suits) defined by $\phi(x,y) = \langle x , y \rangle $ is continuous since

\begin{eqnarray}

|\phi(x,y) – \phi(x’,y’)| &=& |\phi(x,y) – \phi(x,y’) + \phi(x,y’) – \phi(x’,y’)| \\

& \leq & |\phi(x,y) – \phi(x,y’)| + |\phi(x,y’) – \phi(x’,y’)| \\

& \leq & \|x\| \|y-y’\| + \|y’\| \|x-x’\|

\end{eqnarray}

Hence if $\max(\|x-x’\|,\|y-y’\|) \leq 1$, then $|\phi(x,y) – \phi(x’,y’)| \leq \max(\|x\|,\|y\|+1)(\|y-y’\| + \|x-x’\|)$. Now choose $\epsilon>0$, then if $\max(\|x-x’\|,\|y-y’\|) < \frac{1}{2\max(\|x\|,\|y\|+1)}\epsilon$ (and $\leq 1$, of course), you have $|\phi(x,y) – \phi(x’,y’)| < \epsilon$.

Since $\phi$ is continuous, and the map $\eta:S \to \mathbb{R}^n \times \mathbb{R}^n$ defined by $\eta(x) = (f(x),g(x))$ is continuous, it follows that the composition $\phi \circ \eta$ is continuous.

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