Are there non-integer $\nu$, such that the Parabolic Cylinder Function vanishes as $z \to -\infty$? Suppose we have the differential equation
$$
  -f''(x) + \Big( \frac{1}{4} x^2 - \nu - \frac{1}{2} \Big) \, f(x) = 0
  \: , \quad
  \text{where } \nu \in \mathbb{R}
  \: .
$$
The general solution to this equation is $f(x) = a \, D_\nu(x) + b \, D_\nu(-x)$, where $D_\nu$ is the parabolic cylinder function.
If $\nu$ is an non-negative integer, then $D_\nu(x) \propto \exp(-x^2/4)$  is nice and bounded. My question is: are there any $\nu$ other than the non-negative integers, such that
$$
  \lim_{x \to -\infty} D_\nu(x) = 0
$$
(ie. $D_\nu$ vanishes as $x$ goes to negative infinity)?
From trying various different $\nu$ numerically, it seems that the answer is no, however I wasn't able to prove it. The Digital Library of Mathematical Functions includes asymptotic expansions of the parabolic cylinder functions, I just don't know how to use them to prove this.

By http://dlmf.nist.gov/12.9.E3,
$$
D_\nu  (xe^{ \pm \pi i} ) = e^{ \pm \pi \nu i} e^{ - \frac{1}{4}x^2 } x^\nu  \left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right) + \frac{{\sqrt {2\pi } }}{{\Gamma ( - \nu )}}e^{\frac{1}{4}x^2 } x^{ - \nu  - 1} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right)
$$
as $x\to +\infty$. Since the parabolic cylinder function is a single valued entire function, taking the average of the two sides gives
$$
D_\nu  ( - x) = e^{ - \frac{1}{4}x^2 } x^\nu  \left( {\cos (\pi \nu ) +\mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right) + \frac{{\sqrt {2\pi } }}{{\Gamma ( - \nu )}}e^{\frac{1}{4}x^2 } x^{ - \nu  - 1} \left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^2 }}} \right)} \right),
$$
as $x\to +\infty$. This shows that $\lim _{x \to  - \infty } D_\nu  (x) = 0$ if and only if $\nu$ is a non-negative integer.
