# Continous dependence of the minimium of a continuous function over a compact set

Suppose we are working on $\mathbb{C}^n$ and $h = (c_1,\dots,c_k)$ is a unit vector $$|c_1|^2 + \dots + |c_k|^2 = 1$$ Now consider the function for $t\geq0$ $$f(t,h) = \left(\sum_{i=1}^k \left|\sum_{j=0}^n \frac{b_i^{(j)}c_{i+j} e^{-\lambda_i t} t^j}{j!}\right|^2\right) - 1$$ Where $\lambda_i >0$, $b_i^{(j)} = 0$ or $1$ for each $i,j$, also $n<k$ and $c_i=0$ if $i\notin \{1,\dots,k\}$. The function is continuous in both $t$ and $h$. Now since for fixed $h$, $f(t,h)\xrightarrow[t\to\infty]{} = -1$, there is a $t_0$ such that $$f(t,h) <= 0 \quad \forall t\geq t_0$$ So for fixed $h$ with norm one, there is a $t(h)$ defined by $$t(h) = \min\{t_0\,:\;f(t,h) <= 0 \quad \forall t\geq t_0\}$$ (the minimun is reached by continiuty). Now my question is the following, is the function $t(h)$ continuous over the vectors of modulus one given the form of $f$? Notice that $f$ is a really nice function, so i think it's true. This question is not homework, i came with this doing research, so every reference will be appreciated.

• Your function seems to be independent of $h$. Feb 16, 2015 at 7:31
• $f$ depends of $h$ because it depends of the coefficients $c_i$, remember that $h=(c_1,\dots,c_k)$. Feb 16, 2015 at 7:34
• Oh, sorry, you are right. Feb 16, 2015 at 7:34