Difference between revisions of "Aufgaben:Problem 12"

Revision as of 21:50, 15 January 2015

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 the exponential goes to \( 0 \) faster than any polynomial goes to \( \infty \) for \( x \rightarrow \pm \infty: \) $$ \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{-\infty} = \left| p(x) \right| e^{-\frac{1}{2}x^2} \bigg \vert_{\infty} = 0. $$

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

Okay, ladies and gentlemen, listen carefully. Lemma 2 is incredibly boring to prove. We've found a shorter and more general way to do this. I've marked the subparts that divide in two versions separately.

Lemma 2: Let \( f \in \mathcal{S}(\mathbb{R}) \) and \( \lambda (k) \in C^{\infty} \) be a function only 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.

\( \vdash \) : \( \mathcal{S}(\mathbb{R}) \subset L^1( \mathbb{R} ) \)

Indeed, from Series 11, Ex. 1 we know:

$$ \left| f(x) \right| \leq \frac{C}{(1 + \vert x \vert^2)} $$

for some \( C \in \mathbb{R} \) and thus:

$$ \int_{\mathbb{R}} \left| f(x) \right| dx \leq \int_{\mathbb{R}} \left| \frac{C}{(1 + \vert x \vert^2)} \right| dx < \infty $$

Version 1

Using the property

Now, we use an induction argument for \( \left( \frac{d}{dk} \right)^n \hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n f(x) e^{-ikx} dx \)

(well-defined since \( f \in \mathcal{S}(\mathbb{R}) \Rightarrow \hat f(k) \in \mathcal{S}(\mathbb{R}) \subset C^{\infty} \) from a lemma in the script):

\( n = 0 \) is the trivial case.

\( n - 1 \rightarrow n: \)

Let \( h_m \) be an arbitrary zero-sequence.

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

$$ \color{green}{*} \left( \left( \frac{d}{dk} \right)^{j} e^{ - ikx} \right) \bigg \vert_{k + h_m} = \left( p(x)e^{ - ikx} \right) \bigg \vert_{k + h_m} = p(x)e^{ - i(k+h_m)x} = \left( \frac{d}{dk} \right)^{j} e^{ - ikx} e^{ - ih_mx} $$

We then proceed:

$$ \begin{align} \left| \frac{1}{h_m} \left( \frac{d}{dk} \right)^{n-1} f(x) e^{-ikx} \left( e^{-i h_m x} - 1 \right) \right| &= \left| \frac{2}{h_m} f(x) \sin \left( \frac{xh_m}{2} \right) \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| \leq \left| \frac{2}{h_m} f(x) \frac{xh_m}{2} \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| \\ &= \left| x \cdot f(x) \left( \frac{d}{dk} \right)^{n-1} e^{-ikx} \right| = \left| p(x) \cdot f(x) \right| \in L^1 \end{align}$$

where we used \( \vert \sin (x) \vert \leq \vert x \vert \) and \( p(x) \in \mathbb{C}[x] \). Using Lebesgue dominated convergence theorem, the induction proof is concluded.

Version 2:

Using the property from the script: \( \frac{d}{dk} \widehat{f} (k) = (-i) \cdot \widehat{\left( xf \right) }(k) \) we see easily that \( \left( \frac{d}{dk} \right)^n \widehat{f} (k) = \left( -i \right)^n \widehat{ \left( x^n \cdot f \right) } (k) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \left( \frac{d}{dk} \right)^n f(x) e^{-ikx} dx \) since \( \left( \frac{d}{dk} \right)^n e^{-ikx} = (-ix)^n \cdot e^{-ikx} \). If you like you may of course do an induction here but it seems clear enough to me.

Note that \( x^{\alpha} \cdot f \in L^1 \) for any \( \alpha \in \{ 0, ..., n \} \) follows from \( f \in \mathcal{S}(\mathbb{R}) \).

Here the fans of version 1 and version 2 have to work together again.

We now use the generalized Leibniz-rule to show:

$$\begin{align} \left(\frac{d}{dk}\right)^{n} \lambda (k)\int_\mathbb{R} f(x) e^{-ikx}dx &= \sum_{j=0}^{n} \binom{n}{j}\left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \left( \frac{d}{dk} \right)^{n - j} \int_\mathbb{R} f(x) e^{-ikx} dx \\ &= \sum_{j=0}^{n} \binom{n}{j} \left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \int_\mathbb{R} \left( \frac{d}{dk} \right)^{n - j} f(x) e^{-ikx} dx \\ &= \sum_{j=0}^{n} \int_\mathbb{R} \binom{n}{j} \left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \left( \frac{d}{dk} \right)^{n - j} f(x) e^{-ikx} dx \\ &= \int_\mathbb{R} \sum_{j=0}^{n} \binom{n}{j} \left( \left( \frac{d}{dk} \right)^j \lambda (k) \right) \left( \frac{d}{dk} \right)^{n - j} f(x) e^{-ikx} dx \\ &= \int_\mathbb{R} \left( \frac{d}{dk} \right)^n \lambda (k) f(x) e^{-ikx} dx \end{align}$$

And the Lemma is proven. \( \square \)

From this Lemma follows directly for \( f(x) = e^{- \frac{x^2}{2} } \in \mathcal{S}(\mathbb{R}) \) and \( e^{-\frac{k^2}{2}} \in C^{\infty} \)

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

We then finish the proof as follows:

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

Where for the first equality we used Lemma 2 and for the second equation (3.14). \( \square \)

Aternative Solution (not proof-read yet)

claim: \( \hat \Phi_n(k) = (-i)^n\Phi_n(k)\)

proof by induction:

n = 0, using the Fundamental identity in the script (Fourier_SchwartzAdded 95):

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

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

integrating by parts, where the additional term goes to zeros:

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

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

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

taking the differatial outside, using again the mean value theorem and the dominated convergenz theorem. doing it the reverse way:

$$\frac{d}{dk} \int_{\mathbb{R} } e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) i e^{-ikx} dx = \lim\limits_{h \rightarrow 0} \int_{\mathbb{R} } e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) i \frac{e^{-ix(k+h)}-e^{-ixk}}{h} dx$$

(repressing the facotr \((-1)^n\))

$$= \lim\limits_{m \rightarrow \infty} \int_{\mathbb{R} } e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) i \frac{e^{-ix(k+h_m)}-e^{-ixk}}{h_m} dx = \lim\limits_{m \rightarrow \infty} \int_{\mathbb{R} } f_m$$

where \( \lim\limits_{m \rightarrow \infty} h_m = 0 \) an arbitary sequenze going to zero. from the mean value theorem we know that there is a \(\xi\) between 0 and \(h_m\), such that:

$$\frac{e^{-ix(k+h_m)}-e^{-ixk}}{h_m} = -ixe^{-ix(k+\xi)}$$

$$ \Rightarrow |f_m| \leq |x e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big)| = |p(x)e^{-\frac{1}{2}x^2}| \in L_1$$

where p(x) is a polynomial of order n+1. We also know that

$$\lim\limits_{m \rightarrow \infty} f_m= \lim\limits_{m \rightarrow \infty} e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) i\frac{e^{-ix(k+h_m)}-e^{-ixk}}{h_m} dx = e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) i(-ix)e^{-ixk}$$

Thus the dominated Convergenze Theorem applys and we can take the differantial inside.

$$\frac{d}{dk} \int_{\mathbb{R} } e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) i e^{-ikx} dx = \int_{\mathbb{R} } e^{\frac{1}{2}x^2} \Big((\frac{d}{dx})^{n}e^{-x^2}\Big) xe^{-ikx} dx$$

returning to the induction:

$$ (\ast) = i \frac{d}{dk}\hat\Phi_n(k) \ - \ ik\hat\Phi_n(k)$$

using the induction assumtion:

$$= i \frac{d}{dk}\Big( (-i)^n (-1)^n e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^n e^{-k^2} \Big)\Big) \ - \ ik\hat\Phi_n(k)$$

$$= i (-i)^n (-1)^n \Big( k e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^n e^{-k^2} \Big) + e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^{n+1} e^{-k^2} \Big)\Big)\ - \ ik\hat\Phi_n(k)$$

$$= i (-i)^n (-1)^n e^{\frac{1}{2}k^2} \Big( (\frac{d}{dk})^{n+1} e^{-k^2} \Big) = (-i)^{n+1} \Phi_{n+1}(k) $$

\(\square\)

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 \( \eta \rightarrow \pm \infty \).

The result therefore is:

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