Difference between revisions of "Aufgaben:Problem 12"

Revision as of 00:24, 31 December 2014

Part a)

Let \( \Phi_0(x) = e^{-\frac{1}{2}x^2} \) and define \( \Phi_n(x) = (-1)^n e^{\frac{1}{2}x^2}\left(\left(\frac{d}{dx}\right)^n e^{-x^2}\right) \). Show that \( \hat \Phi_n(k) = C\Phi_n(k)\), with \( C = C(n) \in \mathbb{C} \).

Solution

We show this by proving that \( \Phi_n(x) \) is an eigenfunction of the Fourier-transform with eigenvalue \((-i)^n \), i.e. \( \hat \Phi_n(k) = (-i)^n\Phi_n(k).\)

Proof.

We first check the case \( \hat \Phi_0(k) \) (you don't have to know this calculation, we did that in Series 8, ex. 3) where we complete the square in the exponent by using the substitution \( \eta \equiv \frac{x + ik}{\sqrt{2}} \):

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

For the general case:

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

where we partially integrated. The first term of each partial integration is zero since:

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

where \( p(x) = \mathcal{O}(x^j) \) (using the Landau notation) is a (complex) polynomial in \( x \) and \(l, m, j \in \mathbb{N}, \) leading to

$$ \left( \frac{d}{dx} \right)^m e^{\frac{1}{2}x^2 - ikx} \left( \left(\frac{d}{dx}\right)^l e^{-x^2} \right)\bigg \vert_{-\infty}^{\infty} = 0 $$

since $$ \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{-\infty} = \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{\infty} = 0. $$

We then go on like this:

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

where we completed the square in the exponent and then used the following lemma:

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

Proof of Lemma 1:

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 equality we used the Schwarz's theorem. With the induction assumption everything follows. \( \square \)

We then finish the proof as follows:

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

Where for the first equality we used Lemma 4 and for the second equation (3.14). \( \square \)

\( \color{red}{* \: see \: discussion}\)

Lemma 2: \( \frac{d}{dk} \hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \frac{d}{dk} f(x) e^{-ikx} dx \) where \( f(x) = e^{- \frac{1}{2} x^2} \).

Proof of Lemma 2:

We've already shown this in Series 10, Exercise 3.

$$ \sqrt{2\pi} \frac{d}{dk} \hat f(k) = \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{1}{h_n} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh_n} - 1 \right) dx $$

where \(h_n\) is an arbitrary zero-sequence and we used the definition of \( \frac{d}{dk} \) as limit of the difference quotient, the definition of \( \hat f(k) \) and linearity of the integral.

We then use \( \vert \sin(x) \vert \leq \vert x \vert \) (this follows directly from \( \left| \operatorname{sinc}(x) \right| \leq 1 \)) and get:

$$ \begin{align} \left| \frac{1}{h} e^{- \frac{x^2}{2} - ikx} \left( e^{-ixh_n} - 1 \right) \right| &= e^{- \frac{x^2}{2}} \frac{1}{h_n} \left| e^{-ikx} e^{-\frac{ixh_n}{2}} \left( e^{- \frac{ixh_n}{2} } - e^{\frac{ixh_n}{2}} \right) \right| \\ &= e^{- \frac{x^2}{2}} \frac{2}{h_n} \left| \sin \left( \frac{xh_n}{2} \right) \right| \leq \frac{2}{h_n} \left| \frac{xh_n}{2} \right| e^{-\frac{x^2}{2}} \\ &= \vert x \vert e^{-\frac{x^2}{2}} \end{align} $$

and with the Lebesgue dominated convergence theorem the claim holds. \( \square \)

Lemma 3: Let \( f(x) = e^{- \frac{1}{2} x^2} \) and \( \lambda (k) \) be an arbitrary function just depending on k. Then: \( \left( \frac{d}{dk} \right)^{n} \lambda (k) \hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n \lambda (k) f(x) e^{-ikx} dx \).

Proof of Lemma 3:

$$ \begin{align} \frac{d}{dk} \lambda (k) \hat f(k) &= \frac{1}{\sqrt{2\pi}} \frac{d}{dk} \lambda (k) \int_{\mathbb{R}} f(x) e^{-ikx} dx \\ &= \frac{1}{\sqrt{2\pi}} \left( \frac{d}{dk} \lambda (k) \right) \int_{\mathbb{R}} f(x) e^{-ikx} dx + \frac{1}{\sqrt{2\pi}} \lambda (k) \int_{\mathbb{R}} \frac{d}{dk} f(x) e^{-ikx} dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \left( \frac{d}{dk} \lambda (k) \right) f(x) e^{-ikx} + \lambda (k) \frac{d}{dk} f(x) e^{-ikx} dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \frac{d}{dk} \lambda (k) f(x) e^{-ikx} dx \end{align} $$

concluding the prove. \( \square \)

From this follows directly

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

Generalization of Lemma 2:

$$ \left| \frac{1}{h} \left( \frac{d}{dk} \right)^{n-1} f(x) e^{-ikx} \left( e^{-ihx} - 1 \right) \right| = \left| \frac{2}{h} f(x) \sin (\frac{xh}{2}) \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| \leq \left| x f(x) \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| = \left| p(x) f(x) \right| \in L^1 $$

where \( p(x) \in \mathbb{C}[x] \). This holds since \( f \in \mathcal{S}(\mathbb{R}) \). \( \square \)

Generalization of Lemma 3: Use generalized Leibniz rule.

Nice to know, to be checked if necessary to know by heart: Gaussian-function is a Schwatz-fct., Polynomial times Schwartz-function is Schwartz-function (you'll find proof in Dropbox), Schwatz-space subspace of L1 (to be found in Felder-Script).

Part b)

Let \( \chi_{[a,b]} \) be the characteristic function, \( a, b \in \mathbb{R}, a < b \). Show explicitly, i.e. without using Plancherel, that

$$ \Vert \hat \chi_{[a,b]} \Vert_2^2 = b - a. $$

Hint: You may use without proof that

$$ \int_{0}^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2} $$

Solution

Nice to know: Plancherel's theorem states that for any function \( f \in L^2: \Vert \hat f \Vert_{2} = \Vert f \Vert_{2} \) where \( \Vert f \Vert_{2} = \left( \int_{\mathbb{R}} \vert f(x) \vert^2 dx \right)^{\frac{1}{2}} \)

But we show the identity like badasses by direct calculation.

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

Now we use the substitution \( y \equiv \frac{2(x-a)}{b-a} -1 \) to scale the integral to the interval \( \left[ -1, 1 \right] \). Thus:

$$ \begin{align} \hat \chi_{[a,b]}(k) &= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{a+b}{2} \right)}\int_{-1}^{1} e^{-ik \frac{b-a}{2} y} dy \\ &= \frac{(b-a)}{2\sqrt{2\pi}} e^{-ik \left( \frac{a+b}{2} \right)} \left( \frac{-2}{ik(b-a)} e^{-\frac{ik(b-a)}{2}y} \bigg \vert_{-1}^{1} \right) \\ &= \frac{1}{\sqrt{2\pi}} e^{-ik \left( \frac{a+b}{2} \right)} \frac{1}{ik} \left( - e^{-ik\frac{(b-a)}{2}} + e^{ik \frac{b-a}{2}} \right) \\ &= \frac{\sqrt{2}}{k\sqrt{\pi}} e^{-ik \left( \frac{a+b}{2} \right)} \sin \left( \frac{b-a}{2} k \right) \end{align} $$

And so:

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

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

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

It remains to evaluate the integral \( \int_{\mathbb{R} } \frac{\sin^2\left(\eta\right)}{\eta^2} d\eta \).

By using the identity \( \sin^2 \left( x \right) = \frac{1}{2} \left( 1 - \cos \left( 2x \right) \right) \) and partial integration we have

$$ \begin{align} \int_{\mathbb{R} } \frac{\sin^2\left(\eta\right)}{\eta^2} d\eta &= - \frac{1}{2\eta} \left( 1 - \cos \left( 2\eta \right) \right) \Bigg \vert_{- \infty}^{\infty} + \int_{\mathbb{R} } \frac{\sin \left(2\eta \right) }{\eta} d\eta \\ &= 2 \int_{0}^{\infty} \frac{\sin \left(2\eta \right) }{\eta} d\eta \\ &= 2 \int_{0}^{\infty} \frac{\sin\left( \xi \right)}{\xi} d\xi = \pi \end{align}$$

where obviously \( \left| \frac{1}{2\eta} \left( 1 - \cos \left( 2\eta \right) \right) \right| \leq \left| \frac{1}{\eta} \right| \rightarrow 0 \) for \( \eta \rightarrow \infty \).

The result therefore is:

$$ \Vert \hat \chi_{[a,b]}(k) \Vert_2^2 = b-a $$