delta function as a continuous linear map?
mahdiarnt
I've come across a claim that has made me confused. In
http://arxiv.org/abs/quant-ph/9907069 (which is for people like me)
it's claimed on page 17 that the delta function defined on functions
of Schwartz space S as
delta: S -> R, delta_{x_0} f = f(x_0)
is continuous. I think it's true, as implicitly mentioned there as
well, that delta is not continuous on the larger space L^2. However, I
think that the same argument that shows that delta is discontinuous on
V=L^2 also shows that it's discontinuous on V=S as well. Here's the
argument:
If delta: V(=S or L^2) -> R is continuous, then it's a bounded linear
operator. Thus there exists a number M such that
||delta f||_R \leq M ||f||_V,
or
f(x_0)^2 \leq M^2 \int_{-\infty}^{\infty} f(x)^2 dx
But now consider the family of functions
f(x) = exp(-\pi a x^2 / 2) \in S and L^2
and x_0 = 0. The boundedness inequality now reads M^4 \geq a which can
obviously be violated for any fixed M.
Can anyone help please?
G. A. Edgar
<5b79b21d-efa8-4f42...@g8g2000pri.googlegroups.com>,
mahdiarnt <mahd...@yahoo.com> wrote:
> Hi
>
> I've come across a claim that has made me confused. In
> http://arxiv.org/abs/quant-ph/9907069 (which is for people like me)
> it's claimed on page 17 that the delta function defined on functions
> of Schwartz space S as
>
> delta: S -> R, delta_{x_0} f = f(x_0)
>
> is continuous. I think it's true, as implicitly mentioned there as
> well, that delta is not continuous on the larger space L^2. However, I
> think that the same argument that shows that delta is discontinuous on
> V=L^2 also shows that it's discontinuous on V=S as well. Here's the
> argument:
>
> If delta: V(=S or L^2) -> R is continuous, then it's a bounded linear
> operator. Thus there exists a number M such that
>
> ||delta f||_R \leq M ||f||_V,
>
So, you refer to ||f||_S , but what is that? Of course, to determine
whether some function is continuous on S we need to know what is the
topology on S. In fact, this topology is not given by a norm.
> or
>
> f(x_0)^2 \leq M^2 \int_{-\infty}^{\infty} f(x)^2 dx
>
> But now consider the family of functions
>
> f(x) = exp(-\pi a x^2 / 2) \in S and L^2
>
> and x_0 = 0. The boundedness inequality now reads M^4 \geq a which can
> obviously be violated for any fixed M.
>
> Can anyone help please?
>
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Dan Luecking
You have provided a proof that the delta function
is not continuous on L^2. To prove it _is_ continuous
on S, you have to use the topology on S, which is
provided by the fammily of seminorms || f ||_{n,m},
n >= 0, where || f ||_{n,m} is defined to be
\sum_{k,j <= n} sup_{x} (1 + |x|)^j |D^k f(x)|,
(D^k is the n-th derivative.)
To prove continuity, it suffices to find a single value
of n and a constant M such that
|f(x_0)| <= M || f ||_n all f
Since clearly
|f(x_0)| <= sup_{x} |f(x)| = || f ||_0
the delta function is continuous.
Dan
To reply by email, change LookInSig to luecking
Dan Luecking
wrote:
I really meant n,m>0
>
> \sum_{k,j <= n} sup_{x} (1 + |x|)^j |D^k f(x)|,
and I meant \sum_{k <= n, j <= m}
>
>(D^k is the n-th derivative.)
>
>To prove continuity, it suffices to find a single value
>of n and a constant M such that
>
> |f(x_0)| <= M || f ||_n all f
And here || f ||_{n,m}
>
>Since clearly
>
> |f(x_0)| <= sup_{x} |f(x)| = || f ||_0
And here || f ||_{0,0}
mahdiarnt
argument.