2

What I have here is the neighborhood $U := \{z\in\mathbb C : |z| < \varepsilon\}$ of zero in $\mathbb C$ and continuous functions $a,b : U\times [0,1]\to\mathbb C$ such that $|za(t) + b(t)|\ge\delta > 0$ for all $(z,t)\in U\times [0,1]$.

My question: Is the linear span of $\{e^{zt}(za(t)+b(t)) : z\in U\}$ dense in $L^2(0,1)$?

I know this is true for $a=0$ for if $g\in L^2$ is such that $g\perp b(t)e^{zt}$ for all $z\in U$, then $$ 0 = \int_0^1\overline{g(t)}b(t)\sum_{k=0}^\infty\frac{z^kt^k}{k!}\,dt = \sum_{k=0}^\infty\frac{a_k}{k!}z^k, $$ where $a_k = \int_0^1\overline{g(t)}b(t)t^k\,dt$. Since $(a_k)$ is bounded, the series on the rhs defines a holomorphic function on $U$ and hence $a_k = 0$ for all $k$, which means that $bg=0$ and hence $g=0$ since $|b|\ge\delta$. But I don't know how to handle this $z$-dependent function $za+b$.

amsmath
  • 10,633

1 Answers1

0

Your set will not be dense in general.

For instance, take $a=0$, $b=\delta$. In this case you are asking if $\{e^{zt}:\ z\in U\}$ is dense. Consider $g=1_{[0,1/2]}$. Then, writing $z=x+iy$, \begin{align} \|e^{zt}-g\|_2^2&=\int_0^1|e^{zt}-1_{[0,1/2]}(t)|^2\,dt \geq\int_{1/2}^1|e^{zt}-1_{[0,1/2]}(t)|^2\,dt\\ &=\int_{1/2}^1|e^{zt}|^2\,dt =\int_{1/2}^1|e^{xt}|^2\,dt\\ &\geq \int_{1/2}^1|1+xt|^2\,dt\\ &\geq \int_{1/2}^1|1-\varepsilon|^2\,dt\\ &\geq\frac{|1-\varepsilon|^2}2. \end{align}

Martin Argerami
  • 205,756