A quadratic form is continuous on $\mathbb R^n$ 
Prove that the following quadratic function  is continuous in $\mathbb{R}^{n}$,
  $$\textbf{Q}_A(\vec{v})=\vec{v}^{T}\textbf{A}\vec{v}=\sum_{i}^{n}\sum_{j}^{n}\text{a}_{ij}\text{x}_{i}\text{x}_{j}\quad\text{ where }\quad \vec{v}\in\mathbb{R}^{n}\text{ and }\textbf{A}\in\text{Mat}_{n}(\mathbb{R})$$

We know $\sum_{i}^{n}\sum_{j}^{n}\text{a}_{ij}\text{x}_{i}\text{x}_{j}$  is a polynomial with respect to $x_1\dots x_n$  and is continuous on each $x_i\in\mathbb{R}$
 But how to prove $\textbf{Q}_{A}(\vec{v})$  continuous for allt $\vec{v}\in\mathbb{R}^{n}$?
 A: You only need four facts to prove that polynomials are generally continuous:


*

*$+(x,y) = x+y$ is continuous in $\mathbb R^2$

*$\cdot(x,y) = xy$ is continuous in $\mathbb R^2$

*$\cdot_k(x)=kx$ is continuous on $\mathbb R$ for any choice of $k\in R$

*If $f:B\to C$ and $g:A\to B$ are continuous over their respective domains, then so is $f\circ g$

A: First, let $f : \mathbb{R}^n \to \mathbb{R}^n$ and $g: \mathbb{R}^n \to \mathbb{R}^n$ be continuous functions at $a \in \mathbb{R}^n$. Then, the function: $\langle f, g\rangle(p) = \langle f(p), g(p) \rangle$ is continuous at $p$. To prove we use the definition of continuity at $a$. We first have to deduce bound $\langle f,g\rangle$, and we know that we can bound $f(p)-f(a)$ and $g(p)-g(a)$ so we sum and subtract $\langle f(p),g(a)\rangle$ into the norm to make these terms appear and we use billinearity of the inner product. Indeed we have:
$$|\langle f(p),g(p)\rangle-\langle f(a),g(a)\rangle|=|\langle f(p),g(p)\rangle+\langle f(p),g(a)\rangle-\langle f(p),g(a)\rangle-\langle f(a),g(a)\rangle|$$
And by bilinearity of the inner product it is the same as
$$|\langle f(p),g(p)\rangle-\langle f(a),g(a)\rangle|=|\langle f(p),g(p)-g(a)\rangle-\langle f(p)-f(a),g(a)\rangle|$$
Now the triangle inequality gives us:
$$|\langle f(p),g(p)\rangle-\langle f(a),g(a)\rangle|\leq |\langle f(p),g(p)-g(a)\rangle|+|\langle f(p)-f(a),g(a)\rangle|$$
Finally, the Cauchy-Schwarz inequallity gives the most important thing
$$|\langle f(p),g(p)\rangle-\langle f(a),g(a)\rangle|\leq |f(p)||g(p)-g(a)|+|g(a)||f(p)-f(a)|$$
Now, if we make $|f(p)-f(a)| < 1$, but the reverse triangle inequality says that we have the following $||f(p)|-|f(a)||\leq |f(p)-f(a)|$ and so $||f(p)|-|f(a)||<1$ and hence by properties of absolute value, $-1 < |f(p)|-|f(a)| < 1$ and thus $|f(p)| < 1 + |f(a)|$. So, given $\epsilon > 0$ we want to make the following:
$$|f(p)-f(a)| < \min\left\{1, \frac{\epsilon}{2|g(a)|}\right\}$$
$$|g(p)-g(a)| < \frac{\epsilon}{2(|f(a)|+1)}$$
So, by continuity of $f$ at $a$ and of $g$ at $a$ there is $\delta_1 > 0$ that whenever $|p-a|<\delta_1$ the first holds and there is $\delta_2 > 0$ that whenever $|p-a|<\delta_2$ the second holds. If we take $\delta = \min\{\delta_1, \delta_2\}$ both conditions hold, and so, $|p-a|<\delta$ implies that:
$$|\langle f(p),g(p)\rangle-\langle f(a),g(a)\rangle|\leq (1+|f(a)|)|g(p)-g(a)|+|g(a)||f(p)-f(a)|$$
And so:
$$|\langle f(p),g(p)\rangle-\langle f(a),g(a)\rangle|<(1+|f(a)|)\frac{\epsilon}{2(1+|f(a)|)}+|g(a)|\frac{\epsilon}{2|g(a)|}=\epsilon.$$
This proves that the defined function is continuous at $a$. Now, linear transformations are continuous and the quadratic form is just $Q(v)=\langle Av, v\rangle$ where the inner product is the euclidean one. Since both multiplication by a matrix and identity are continuous, the previous theorem grants that the desired function is continuous as expected.
