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.

  • $\begingroup$ Your function seems to be independent of $h$. $\endgroup$
    – PhoemueX
    Feb 16, 2015 at 7:31
  • 1
    $\begingroup$ $f$ depends of $h$ because it depends of the coefficients $c_i$, remember that $h=(c_1,\dots,c_k)$. $\endgroup$
    – Héctor
    Feb 16, 2015 at 7:34
  • $\begingroup$ Oh, sorry, you are right. $\endgroup$
    – PhoemueX
    Feb 16, 2015 at 7:34


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