Fourier Transform Convolution and Approximate Identities — Examples

The interval indicator is integrable with norm $b-a$. For $a<b$ and $\xi \mathrm{\ne }0$, F2 gives the antiderivative $-e^{-2\pi ix\xi }\mathrm{/}(2\pi i\xi )$. F1 at a and b gives the displayed quotient. If $a=b$ the indicator is null almost everywhere and both the numerator and transform vanish.

At zero frequency the integral is the interval length $b-a$; F3 shows this is the continuous extension of the quotient. For the symmetric interval its numerator is $e^{\pi i\xi }-e^{-\pi i\xi }=2i\sin (\pi \xi )$ by F2, giving the sinc formula and value one. Countable choice is inherited from the interval Lebesgue measure and FTC bridge.

Write $g_t=g_1\circ (\sqrt{t}I)$. F2 and F1 give ${\widehat{g}}_t(\xi )=t^{-n\mathrm{/}2}{\widehat{g}}_1(\xi \mathrm{/}\sqrt{t})=t^{-n\mathrm{/}2}{e}^{-\pi \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{/}t}$. Thus the spatial scale is $t^{-1\mathrm{/}2}$, while the frequency scale is $t^{1\mathrm{/}2}$.

At zero frequency the formula gives $\int g_t=t^{-n\mathrm{/}2}$. Multiplying both sides by $t^{n\mathrm{/}2}$ gives the mass-one formula. For t=1 the Gaussian is unchanged by Fourier transform; in n coordinates the product of n one-dimensional values gives the factor $t^{-n\mathrm{/}2}$. Countable choice is inherited from F1 and F2.

Put $q_a(x)=e^{-2\pi a\mathrm{\mid }x\mathrm{\mid }}$. It is integrable with integral $1\mathrm{/}(\pi a)$, by F1 on finite half-intervals and F2 for the positive exponential tails. Integration on $[0,R]$ gives $[1-e^{-2\pi (a+i\xi )R}]\mathrm{/}[2\pi (a+i\xi )]$. The omitted absolute tail is $e^{-2\pi aR}\mathrm{/}(2\pi a)$, so the half-line transform is $1\mathrm{/}[2\pi (a+i\xi )]$. Reflecting the negative half gives $1\mathrm{/}[2\pi (a-i\xi )]$. Their sum is $a\mathrm{/}[\pi (a^2+{\xi }^2)]=P_a(\xi )$.

The rational P_a is bounded on a compact core and bounded by $a\mathrm{/}(\pi x^2)$ for $\mathrm{\mid }x\mathrm{\mid }\ge 1$, hence integrable by the elementary convergent inverse-square improper tail and F2. F3 applied to q_a gives ${\widehat{P}}_a(-x)=q_a(x)$ almost everywhere. Both sides are continuous, so equality holds everywhere (a nonzero continuous difference cannot vanish a.e. on an interval). Evenness then gives the stated transform and, at zero, $\int P_a=1$.

In the Abel integral, inserting $\widehat{f}$ gives a double absolute bound $\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }{q}_a{\mathrm{\parallel }}_1<\mathrm{\infty }$. F4 exchanges the integrals and step 1.1 identifies the inner inverse integral with $P_a(x-y)$, since P_a is even. Thus the mean is $f\ast P_a$, absolutely at every x since P_a is bounded. Finally $P_a(y)=a^{-1}{P}_1(y\mathrm{/}a)$ and $\mathrm{\Phi }(r)=1\mathrm{/}[\pi (1+r^2)]$ is bounded, decreasing and integrable with radial mass one. F5 gives the asserted Lebesgue-point limit. All integral and pointwise suppliers carry the stated countable choice.

The integral $(I\ast I)(x)$ is the length of $[-1\mathrm{/}2,1\mathrm{/}2]\cap [x-1\mathrm{/}2,x+1\mathrm{/}2]$. For $0\le x\le 1$ this length is $1-x$; for $-1\le x\le 0$ it is $1+x$; for $\mathrm{\mid }x\mathrm{\mid }>1$ the intervals are disjoint. At $\mathrm{\mid }x\mathrm{\mid }=1$ the intersection is a null singleton. Thus $I\ast I=T$ everywhere.

By F1 and F2, $\widehat{T}={\widehat{I}}^2$ gives the displayed squared sinc. At zero the value is one, also equal to ${\int }_{-1}^1(1-\mathrm{\mid }x\mathrm{\mid })dx=2[x-x^2\mathrm{/}2{]}_0^1=1$. Countable choice is inherited from the indicator and convolution suppliers.

Take the f in F1, whose integral and norm are one. For integer $k\ge 1$ and $\xi \in [k+1\mathrm{/}6,k+5\mathrm{/}6]$, $\mathrm{\mid }\sin (\pi \xi )\mathrm{\mid }\ge 1\mathrm{/}2$. Consequently $\mathrm{\mid }\widehat{f}(\xi )\mathrm{\mid }\ge 1\mathrm{/}[2\pi (k+1)]$ on an interval of length $2\mathrm{/}3$, and its integral there is at least $1\mathrm{/}[3\pi (k+1)]$.

These intervals are disjoint. The sum of their lower bounds diverges: the block $2^m\le k<2^{m+1}$ contributes at least $2^m\mathrm{/}[3\pi 2^{m+1}]=1\mathrm{/}(6\pi )$. Hence $\int \mathrm{\mid }\widehat{f}\mathrm{\mid }=\mathrm{\infty }$, despite $\mathrm{\parallel }f{\mathrm{\parallel }}_1=1$. The failure concerns absolute integrability, not continuity or decay of the transform. Countable choice is inherited from F1 and Lebesgue interval measure.

F1 and F2 give $\int \mathrm{\mid }f\mathrm{\mid }=0$, so f represents the zero integrable class. F3 gives $\widehat{f}(\xi )=0$ for every frequency, and hence the transform is integrable too.

The inverse integral at zero is $\int \widehat{f}(\xi )d\xi =0$, whereas $f(0)=1$. Thus the claimed every-point assertion fails. In fact the mean oscillation about the chosen value f(0) equals one on every centered interval, so zero is not a Lebesgue point with that value. This respects the actual almost-everywhere inversion theorem. Countable choice is inherited from F1.

Set ${\xi }_0=0$. For $k\ge 1$ let ${\xi }_k$ be the least positive integer greater than ${\xi }_{k-1}+k$ for which $r({\xi }_k)<2^{-2k}$. Such integers exist by the assumed limit. This recursion is explicit and uses no choice selection. The series $f(x)={\sum }_{k\ge 1}{2}^{-k}{e}^{2\pi i{\xi }_{k}x}g(x)$ converges absolutely at each x with modulus at most g(x); its measurable limit belongs to $L^1$ by F1.

At each frequency, F3 applied to the partial sums times the unit-modulus Fourier factor, dominated by g, gives $\widehat{f}(\xi )={\sum }_{k\ge 1}{2}^{-k}{e}^{-\pi (\xi -{\xi }_k{)}^2}$ by F1 and F2. All summands are nonnegative. Thus $\mathrm{\mid }\widehat{f}({\xi }_k)\mathrm{\mid }\ge 2^{-k}$ and $\mathrm{\mid }\widehat{f}({\xi }_k)\mathrm{\mid }\mathrm{/}r({\xi }_k)>2^k\to \mathrm{\infty }$, while ${\xi }_k\to \mathrm{\infty }$. This refutes every proposed rate, even allowing an f-dependent big-O constant. Countable choice is inherited only from the Gaussian and modulation suppliers. This explicit construction is local and is not attributed to a source theorem.