Are eigenvalues continuous functions of a matrix function? Consider a matrix-valued function $A(t):\mathbb{R}_+ \to \mathbb{R}^{n \times n}$. Suppose that each element $a_{ij}(t)$ is a smooth function with bounded derivative, e.g. $a_{ij}(t)=A_{ij}\sin(\omega_{ij} t)$.
Define $f(t)$ as a minimum by absolute value eigenvalue of $A(t)$: $$f(t) = \min_i |\lambda_i\{A(t)\}|.$$
Is it true that $f(t)$ is continuous differentiable and has bounded derivative? If so, then how it can be proven, or in which book/paper it can be found?
Update
Ok, $f(t)$ is continuous, but probably not differentiable. I have revised my main problem and I see that I can alleviate the question. So, now I need $f(t)$ to be Lipschitz. 
Actually, what I really need it to show that if for some $t_0 \in [t_a,t_b]$ all eignvalus of $A(t)$ are nonzero, then $$\int_{t_a}^{t_b}\left(\det\{A(s)\}\right)^2ds>0.$$ My intention was to use $\left(\det\{A(t)\}\right)^2=\prod_i^n|\lambda_i\{A(t)\}|^2\ge \left(\min_i |\lambda_i\{A(t)\}|\right)^{2n} = f(t)^{2n}$.
Note also that all $a_{ij}(t)$ are bounded.
 A: Differentiable is hopeless.  The minimum and absolute value are not differentiable, so this is a lot to expect.  Example:
$$  \begin{pmatrix} \sin t & 0 \\ 0 & \cos t \end{pmatrix}  \text{.}  $$
We plot the minimum eigenvalue for this matrix

as well as the absolute values of the eigenvalues (tracking which is which by color).

Update : 
From Eigenvalues of matrix with entries that are continuous functions , the eigenvalues of your matrix are continuous functions of $t$.  Consequently, since the $\lambda_i\{A(t_0)\} \neq 0$ simultaneously, your product, $\prod_{i=1}^n |\lambda_i\{A(t)\}|^2$ is positive on a neighbourhood of $t_0$.
Alternatively, the determinant is a continuous function of the entries, which are continuous functions, so the determinant is a continuous function of $t$.  Since the determinant is the product of the eigenvalues, which are simultaneously nonzero at $t_0$, the determinant is nonzero on a neighbourhood of $t_0$.
Either way, the integral you write has an everywhere nonnegative integrand and we have shown that it is positive on an interval, so we have shown that the integral is positive.
Update 2 : 
I have to go do other work.  But this last version seems likely.  The determinant is a quasilinear (no variable appears with power $ >1 $ in any term) polynomial in the entries of the matrix, so the derivative of the determinant will be some magnificently huge polynomial in the zeroeth and first derivatives of the matrix entries...
A: No. Absolute value is not differentiable at 0. For example, take $n=1$ and $A (t)=t$. 
