Show multivariable quadratic function is continuous? I want to show that $f(x)=x^TAx+c^Tx$ where $f: \mathbb{R}^n \rightarrow \mathbb{R}$, $A$ is an $n \times n$ matrix, and $c$ is in $\mathbb{R}^n$ is a continuous function.
Using the definition we have:
$$
\forall \varepsilon > 0 : \exists \delta > 0: \forall x \in \mathbb{R}^n \,\,\,\, \|x-x_c\| < \delta \implies |f(x)-f(x_c)| < \varepsilon.
$$
where $\| \cdot\|$ is Euclidean norm.
So using triangle inequality we have
$$
|f(x)-f(x_c)|=|x^TAx+c^Tx-x_c^TAx_c-c^Tx_c| \leq |x^TAx-x_c^TAx_c|+|c^Tx-c^Tx_c|
$$
Using Cauchy-Schwarz we have
$$
|x^TAx-x_c^TAx_c|+|c^Tx-c^Tx_c| \leq |x^TAx-x_c^TAx_c|+ \|c\| \|x-x_c\|
$$
To have $|x^TAx-x_c^TAx_c|$ in terms of $\|x-x_c\|$ we need to do something which I do not know?
Maybe
$$
|x^TAx-x_c^TAx_c|=|x^TAx-x^TAx_c+x^TAx_c-x_c^TAx_c|
$$
Please complete this answer not starting new one.
 A: Writing
$\Vert x^TAx-x_c^TAx_c \Vert =\Vert x^TAx-x^TAx_c+x^TAx_c-x_c^TAx_c \Vert \tag 1$
is certainly on the right track, and indeed very close to the station, since
$\Vert x^TAx-x^TAx_c+x^TAx_c-x_c^TAx_c \Vert \le \Vert x^TAx-x^TAx_c \Vert + \Vert x^TAx_c-x_c^TAx_c \Vert$
$= \Vert x^TA(x - x_c) \Vert + \Vert (x^T - x_c^T)Ax_c \Vert \le \Vert x^T A \Vert \Vert x - x_c \Vert + \Vert x^T - x_c^T \Vert \Vert Ax_c \Vert; \tag 2$
now $x_c$ is fixed, hence so is $\Vert A x_c \Vert$; and also
$\Vert x^T A \Vert \le \Vert x^T \Vert \Vert A \Vert; \tag 3$
if now
$\Vert x - x_c \Vert < \delta, \tag 4$
then
$\vert \Vert x \Vert - \Vert x_c \Vert \vert \le \Vert x - x_c \Vert < \delta, \tag 5$
which implies that
$\Vert x \Vert < \Vert x_c \Vert + \delta; \tag 6$
furthermore, in the Euclidean norm
$\Vert x^T \Vert = \Vert x \Vert, \tag 7$
and thus (3) becomes
$\Vert x^T A \Vert \le \Vert x \Vert \Vert A \Vert < (\Vert x_c \Vert + \delta) \Vert A \Vert; \tag 8$
we may now assemble (1), (2) and (8) together and write
$\Vert x^TAx-x_c^TAx_c \Vert \le (\Vert x_c \Vert + \delta) \Vert A \Vert  \Vert x - x_c \Vert +  \Vert x - x_c \Vert \Vert Ax_c \Vert$
$= ((\Vert x_c \Vert + \delta) \Vert A \Vert  + \Vert Ax_c \Vert) \Vert x - x_c \Vert, \tag 9$
whenever $x$ is such that (4) binds.
Then to attain
$\Vert x^TAx-x_c^TAx_c \Vert < \epsilon, \tag{10}$
we need only ensure
$\Vert x - x_c \Vert < \min \{((\Vert x_c \Vert + \delta) \Vert A \Vert  + \Vert Ax_c \Vert)^{-1} \epsilon, \delta \}, \tag{11}$
and we have established the requisite continuity via what is essentially an $\epsilon$-$\delta$ argument.
