Difference between revisions of "Aufgaben:Problem 12"

Revision as of 10:16, 27 December 2014

Draft: 3 a)

\(\vdash : \Phi_n(x) = (-1)^n e^{\frac{1}{2}x^2} \Big(\frac{d}{dx}\Big)^n e^{-x^2} \) is an eigenfunction of the Fourier-transform with the eigenvalue \((-i)^n\).

Proof.

We first check the case

$$ \hat \Phi_0(k) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R} } e^{-\frac{1}{2}x^2} e^{-ikx} dx $$

with the substitution

$$ \eta \equiv \frac{x + ik}{\sqrt{2}} $$

we get

$$ \hat \Phi_0(k) = \frac{\sqrt{2}}{\sqrt{2\pi}} \int_{\mathbb{R} } e^{-\eta^2 - \frac{k^2}{2}} d\eta = e^{-\frac{1}{2} k^2} = \Phi_0(k) \tag{3.14} $$

By direct calculation:

$$ \sqrt{2\pi} \hat \Phi_n(k) = \int_{\mathbb{R} } (-1)^n e^{\frac{1}{2}x^2} \Big( (\frac{d}{dx})^n e^{-x^2} \Big) e^{-ikx} dx = \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}x^2 - ikx} dx $$

where we partially integrated. The equation makes sense since

$$ | \Big( \frac{d}{dx} \Big)^m e^{\frac{1}{2}x^2 - ikx} \Big( (\frac{d}{dx})^l e^{-x^2} \Big) | = f(x) e^{-\frac{1}{2}x^2} $$

with

$$ f(x) = O(n^p) $$

for

$$ n, l, p \in \mathbb{N} $$.

So we have:

$$ \sqrt{2\pi} \Phi_n(k) = e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx = i^n e^{\frac{1}{2}k^2} \int_{\mathbb{R} } e^{-x^2} \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} dx $$

Lemma: \( \forall n \in \mathbb{Z}_{\geqslant 0}: i^n \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} = \Big( \frac{d}{dx} \Big)^n e^{\frac{1}{2}(x - ik)^2} \)

Proof of Lemma:

By induction.

\( n = 0 \) : \( i^0 e^{\frac{1}{2}(x - ik)^2} = e^{\frac{1}{2}(x - ik)^2} \) is the trivial case.

\( n-1 \rightarrow n \) :

$$ i^n \Big( \frac{d}{dk} \Big)^n e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \Big( \frac{d}{dk} \Big)^{n-1} i (-i) (x-ik) e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \Big( \frac{d}{dk} \Big)^{n-1} (x-ik) e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \Big( \frac{d}{dk} \Big)^{n - 1} \frac{d}{dx} e^{\frac{1}{2}(x - ik)^2} = i^{n-1} \frac{d}{dx} \Big( \frac{d}{dk} \Big)^{n - 1} e^{\frac{1}{2}(x - ik)^2} $$

where for the last equation we used the Schwarz's theorem. With the induction assumption everything follows. \( \square \)

The rest is a piece of cake:

$$ \sqrt{2\pi} \hat \Phi_n(k) = i^n e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n \int_{\mathbb{R} } e^{-x^2 + \frac{1}{2}(x - ikx)^2} dx = i^n \sqrt{2\pi} e^{\frac{1}{2}k^2} \Big( \frac{d}{dk} \Big)^n e^{-k^2} = (-i)^n \sqrt{2\pi} \Phi_n(k) $$

(for the second last equality we've used equation (3.14) ) . $$\square$$

Draft: 3 b) (wird noch alles schön dargestellt)

$$ \vdash : \Vert \hat \chi_{[a,b]} \Vert_2^2 = b - a $$ for $$ a, b \in \mathbb{R} $$.

Proof.

We show the identity without making use of Plancherel's theorem.

We have: $$ \hat \chi_{[a,b]} = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \chi_{[a,b]}(x) e^{-ikx} dx = \frac{1}{\sqrt{2\pi}} \int_{a}^b e^{-ikx} dx$$

Now, let: $$ x \equiv \frac{2(y-a)}{b-a} -1 $$

$$ \hat \chi_{[a,b]} = \frac{2}{(b-a)\sqrt{2\pi}} e^{-\frac{1}{2} ik(b + a)}\int_{-1}^{1} e^{-ik \frac{b-a}{2} y} dy = \frac{2}{\sqrt{2\pi}} e^{-\frac{1}{2} ik(b + a)} \frac{sin(\frac{b-a}{2}k)}{k} $$

And thus:

$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \int_{\mathbb{R} } |\hat \chi_{[a,b]}(k)|^2 dk = \frac{2}{\pi} \int_{\mathbb{R} } \frac{sin^2(\frac{b-a}{2}k)}{k^2} dk $$

With the substitution $$ \eta \equiv \frac{b-a}{2}k $$ we get:

$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = \frac{(b-a)^2}{(b - a) \pi} \int_{\mathbb{R} } \frac{sin^2(\eta)}{\eta^2} d\eta $$

It remains to evaluate the integral $$ I(\alpha) = \int_{\mathbb{R} } \frac{\sin^2(\alpha \eta)}{\eta^2} d\eta $$ at $$ \alpha = 1$$.

$$ \frac{dI}{d\alpha} = 2 \int_{0}^{\infty} \frac{2 \eta \sin(\alpha \eta) \cos(\alpha \eta) }{\eta^2} $$ and with the identity $$ \sin(x) \cos(y) = \frac{1}{2}\Big(\sin (x-y) + \sin (x+y)\Big) $$ it follows that:

$$ \frac{dI}{d\alpha} = \pi $$.

Solving this equation is a thing we learned a long time ago:

$$ I(\alpha) = \pi \alpha + c $$ and since $$ I(0) = 0 $$ :

$$ I(\alpha) = \pi \alpha $$.

As $$ I(1) = \pi $$ the claim is proven. $$\square$$