Aufgaben:Problem 12

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

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

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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.

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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 \( f \in \mathcal{S}(\mathbb{R}) \).

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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:

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