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Introduction to Linear Algebra with Mathematica
Glossary
Preface
We denote by 𝔏¹(ℝ) the vector space of Lebesgue integrable functions on the real line. Its elements are equivalent classes of functions that are equal almost everywhere. The space 𝔏¹(ℝ) or 𝔏¹ is the vector space of Lebesgue integrable functions, defined "up to equality almost everywhere," with the norm
In this section, we define the Fourier transform of integrable functions from 𝔏¹(ℝ) and derive its main properties.
Fourier transform in 𝔏¹
- ℱ(f + g) = ℱ(f) + ℱ(g).
- For any complex number α, \[ ℱ\left( \alpha\, f \right) = \alpha ℱ\left( f \right) \]
- If f✶ (also denoted by \( \displaystyle \ \overline{f} \) ) is complex conjugate of f, then \[ ℱ\left( f^{\ast}\right) = \hat{f}(-\xi )^{\ast} = \overline{\hat{f}(-\xi )} . \]
- Let us denote the shift of function f by fy(x) = f(x + y) ror y ∈ ℝ. Then \[ \hat{f}_y (\xi ) = \hat{f}(\xi )\,e^{-{\bf j}\xi y} . \]
- \( \displaystyle \quad \left\vert \hat{f} (\xi ) \right\vert \leqslant \int \left\vert f(x) \right\vert {\text d}x = \| f \|_1 . \)
- For positive λ, let φ(x) = λ f(λx), then \[ \hat{\varphi}(\xi ) = f \left( \frac{\xi}{\lambda} \right) . \]
- Linearity: for f,g ∈ 𝔏¹(ℝ), we have \[ \hat {f+g}(\xi ) =\int _{\mathbb{R}}(f(x)+g(x))e^{-{\bf j}\xi x}\, {\text d}x = \int_{\mathbb{R}}f(x)\.e^{-{\bf j}\xi x}\, {\text d}x + \int_{\mathbb{R}}g(x)e^{-{\bf j}\xi x}\, {\text d}x = \hat {f}(\xi ) +\hat {g}(\xi ). \]
- Homogeneity: for α ∈ ℂ and f ∈ 𝔏¹(ℝ), we have \[ \hat {\alpha f}(\xi )=\int _{\mathbb{R}}\alpha f(x)e^{-i\xi x}\, dx=\alpha \int _{\mathbb{R}}f(x)e^{-i\xi x}\, dx=\alpha \hat {f}(\xi ). \]
- Conjugation: Let \( \displaystyle \quad f^*(x)=\overline{f(x)}. \ \) Then \[ \hat {f^*}(\xi )=\int _{\mathbb{R}}\overline{f(x)}\, e^{-i\xi x}\, dx. \] Take complex conjugate of ℱ[f](-ξ): \[ \hat {f}(-\xi )=\int _{\mathbb{R}}f(x)e^{-i(-\xi )x}\, dx=\int _{\mathbb{R}}f(x)e^{i\xi x}\, dx. \] Thus, \[ \overline{\hat {f}(-\xi )}=\overline{\int _{\mathbb{R}}f(x)e^{i\xi x}\, dx}=\int _{\mathbb{R}}\overline{f(x)e^{i\xi x}}\, dx=\int _{\mathbb{R}}\overline{f(x)}\, e^{-i\xi x}\, dx=\hat {f^*}(\xi ). \] So \[ \hat {f^*}(\xi )=\overline{\hat {f}(-\xi )}. \]
- Translation: \( \displaystyle \quad \hat{f_y}(\xi ) =\hat{f}(\xi )\,e^{-{\bf j}\xi y} \quad \) for fy(x) = f(x+y). Then \[ \hat {f_y}(\xi )=\int _{\mathbb{R}}f(x+y)e^{-i\xi x}\, dx. \] Use the change of variable u = x+y, so x = u-y, dx = du: \[ \hat {f_y}(\xi )=\int _{\mathbb{R}}f(u)e^{-i\xi (u-y)}\, du=e^{i\xi y}\int _{\mathbb{R}}f(u)e^{-i\xi u}\, du=e^{i\xi y}\hat {f}(\xi ). \] If your convention uses \( \displaystyle \quad e^{-{\bf j}\xi x} \quad \) in the transform, the factor is \( \displaystyle \quad e^{{\bf j}\xi y}. \ \) If the statement you were given has \( \displaystyle \quad e^{-{\bf j}\xi y}, \ \) then the underlying sign convention in the exponential is the opposite. With the above definition, the correct factor is ejξy. If your notes use a slightly different Fourier transform convention (e.g., \( \displaystyle \quad e^{+{\bf j}\xi x} \quad \) or 2π factors), the proofs are identical, but the signs and scaling factors adjust accordingly.
-
Linearity: \( \displaystyle \ \hat {f+g} =\hat {f}+\hat {g} . \)
Take \[ f(x) = e^{-|x|},\qquad g(x) =\mathbf{1}_{[-1,1]}(x). \] Then \begin{align*} \hat {f}(\xi ) &= \int_{\mathbb{R}} e^{-|x|}e^{-{\bf j}\xi x}\, {\text d}x = \frac{2}{1+\xi ^2}. \\ \hat {g}(\xi ) &= \int _{-1}^1 e^{-{\bf j}\xi x}\, {\text d}x = \frac{2\sin \xi }{\xi }. \end{align*}
Integrate[Exp[-Abs[x]]*Exp[-I*x*t], {x, -Infinity, Infinity}, Assumptions -> {t \[Element] Reals}]2/((-I + t) (I + t))Integrate[Exp[-I*x*t], {x, -1, 1}, Assumptions -> {t \[Element] Reals}](2 Sin[t])/tNow compute ℱ[ f + g ]: \[ \hat {f+g}(\xi ) =\int_{\mathbb{R}}(e^{-|x|} +\mathbf{1}_{[-1,1]}(x))\, e^{-{\bf j}\xi x}\, {\text d}x =\frac{2}{1+\xi^2} + \frac{2\sin \xi }{\xi }. \] This matches ℱ[ f ](ξ) + ℱ[ g ](ξ). -
Scalar multiplication: ℱ[ α f ] = α ℱ[ f ](ξ).
Let \( \displaystyle \quad f(x) =e^{-x^2} , \ \) then the Gaussian transform is known: \[ \hat {f}(\xi ) = \sqrt{\pi }\, e^{-\xi ^2/4}. \]
Integrate[Exp[-x^2]*Exp[-I*x*t], {x, -Infinity, Infinity}, Assumptions -> {t \[Element] Reals}]E^(-(t^2/4)) Sqrt[\[Pi]]Then \[ \hat {3{\bf j}\,f}(\xi ) =3{\bf j}\,\sqrt{\pi }\, e^{-\xi ^2/4} = 3{\bf j}\, \hat {f}(\xi ). \] -
Conjugation: \( \displaystyle \quad \hat {f^*}(\xi ) =\overline{\hat {f}(-\xi )} . \)
Let \[ f(x) = e^{{\bf j}\,x}\mathbf{1}_{[0,1]}(x). \] Then \[ f^*(x)=\overline{e^{ix}}\mathbf{1}_{[0,1]}(x)=e^{-ix}\mathbf{1_{\mathnormal{[0,1]}}}(x). \] Compute: \[ \hat {f}(\xi )=\int _0^1e^{{\bf j}x}e^{-{\bf j}\xi x}\, {\text d}x =\int_0^1 e^{-{\bf j}(\xi -1)x}\, {\text d}x = \frac{1-e^{-{\bf j}(\xi -1)}}{{\bf j}(\xi -1)}. \]
Integrate[Exp[I*x]*Exp[-I*x*t], {x, 0, 1}, Assumptions -> {t \[Element] Reals}](I (-1 + E^(-I (-1 + t))))/(-1 + t)Similarly, \[ \hat {f^*}(\xi ) = \int _0^1 e^{-{\bf j}\,x} e^{-{\bf j}\xi x}\, {\text d}x =\frac{1-e^{-{\bf j}(\xi +1)}}{{\bf j}(\xi +1)}. \]Integrate[Exp[-I*x]*Exp[-I*x*t], {x, 0, 1}, Assumptions -> {t \[Element] Reals}](I (-1 + E^(-I (1 + t))))/(1 + t)Now we check the identity: \[ \overline{\hat {f}(-\xi )}=\overline{\frac{1-e^{-i(-\xi -1)}}{i(-\xi -1)}}=\frac{1-e^{-i(\xi +1)}}{i(\xi +1)}=\hat {f^*}(\xi ). \] -
Translation: \( \displaystyle \quad \hat {f_y}(\xi ) = \hat {f}(\xi )e^{{\bf j}\xi y} . \)
Let f(x) = e-|x| again, and choose y = 2. Then \[ f_2(x)=f(x+2)=e^{-|x+2|}. \] We know ℱ[ f ](ξ) = 2/(1 + ξ²).
Integrate[Exp[-Abs[x]]*Exp[-I*x*t], {x, -Infinity, Infinity}, Assumptions -> {t \[Element] Reals}]2/((-I + t) (I + t))The theorem predicts: \[ \hat {f_2}(\xi )=\frac{2}{1+\xi ^2}\, e^{{\bf j}2\xi }. \] You can verify this directly by substituting u = x + 2 in the integral.Integrate[Exp[-Abs[x + 2]]*Exp[-I*x*t], {x, -Infinity, Infinity}, Assumptions -> {t \[Element] Reals}](2 E^(2 I t))/((-I + t) (I + t)) -
𝔏¹-bound: \( \displaystyle \quad |\hat {f}(\xi )|\leq \| f\|_1 . \)
Take \[ f(x) =\frac{1}{1+x^2}. \] Then \[ \| f\|_1 =\int _{\mathbb{R}}\frac{{\text d}x}{1+x^2} =\pi . \]
Integrate[1/(1 + x^2), {x, -Infinity, Infinity}]\[Pi]The Fourier transform is known: \[ \hat {f}(\xi ) =\pi e^{-|\xi |}. \]Integrate[Exp[-I*x*t]/(1 + x^2), {x, -Infinity, Infinity}, Assumptions -> {t \[Element] Reals}]E^-Abs[t] \[Pi]Check the bound: \[ \pi = \sup \left\{ \pi\,e^{-|\xi |} \right\} = \| \hat{f} \|_{\infty} \leqslant \| f \|_1 = \pi . \]This shows the inequality is sharp at ξ = 0.
-
Scaling: if φ(x) = λ f(λ x), then \( \displaystyle \quad \hat {\varphi }(\xi ) =\hat {f}(\xi /\lambda ) . \)
Let \( \displaystyle \quad f(x) =\mathbf{1}_{[-1,1]}(x). \quad \) Then \[ \hat {f}(\xi )=\frac{2\sin \xi }{\xi }. \] Choose \[ \varphi (x)=3\, f(3x)=3\, \mathbf{1}_{[-1/3,\, 1/3]}(x). \] The theorem predicts: \[ \hat {\varphi }(\xi )=\hat {f}(\xi /3)=\frac{2\sin (\xi /3)}{\xi /3}. \] You can also compute directly: \[ \hat {\varphi }(\xi )=\int _{-1/3}^{1/3}3e^{-i\xi x}\, dx=3\cdot \frac{2\sin (\xi /3)}{\xi }=\frac{2\sin (\xi /3)}{\xi /3}, \] which matches perfectly.
Now we prove that there is no function q(x) ∈ 𝔏¹ such that ℱ[q] = Q.
We consider the following limit: \[ \lim_{N\to\infty} \int_e^N \frac{Q(\xi )}{\xi}\,{\text d}\xi = \lim_{N\to\infty} \int_e^N \frac{{\text d}\xi}{\xi\,\ln\xi} = \lim_{N\to\infty} \ln \ln N = \infty . \] This result shows that integral \[ \int_e^{\infty} \frac{Q(\xi )}{\xi}\,{\text d}\xi \qquad \mbox{diverges} . \]
Suppose there exists function q(x) ∈ 𝔏¹ such that ℱ[q] = Q, \[ Q(\xi ) = ℱ[q] = \int_{-\infty}^{\infty} q(x)\, e^{-{\bf j}\xi x} {\text d} x . \] Since function Q(ξ) is odd, we have \[ Q(\xi ) = - \int_{-\infty}^{\infty} q(x)\, e^{{\bf j}\xi x} {\text d} x . \] Adding these two equations, we get \[ Q(\xi ) = \int_{-\infty}^{\infty} q(x) \left[ e^{-{\bf j}\xi x} - e^{{\bf j}\xi x} \right] {\text d} x = -2{\bf j} \int_{-\infty}^{\infty} q(x) \,\sin (\xi x)\,{\text d} x . \] We break it into \begin{align*} Q(\xi ) &= - 2{\bf j}\int_{-\infty}^{0} q(x) \,\sin (\xi x)\,{\text d} x - 2{\bf j}\int_{0}^{\infty} q(x) \,\sin (\xi x)\,{\text d} x \\ &= - 2{\bf j}\int_{0}^{\infty} \left[ q(x) - q(-x) \right] \sin (\xi x)\,{\text d} x \\ &= \int_{0}^{\infty} q_0 (x)\,\sin (\xi x)\,{\text d} x , \qquad q_0 = -2{\bf j} \left[ q(x) - q(-x) \right] . \end{align*} It is clear that q₀ ∈ 𝔏¹([0,∞)). Then we have \begin{align*} \int_e^N \frac{Q(\xi )}{\xi}\,{\text d}\xi &= \int_e^N \frac{{\text d}\xi}{\xi} \int_{0}^{\infty} q_0 (x)\,\sin (\xi x)\,{\text d} x \\ &= \int_{0}^{\infty} q_0 (x)\,{\text d}x \int_e^N \frac{\sin (\xi x)}{\xi}\,{\text d}\xi . \end{align*} Fubini's theorem guarantees that exchange of order of integration is valid since function q₀sin(ξx)/ξ is absolutely integrable. The inner integral \[ \int_e^N \frac{\sin (\xi x)}{\xi}\,{\text d}\xi \qquad\mbox{converges as } N\to\infty . \] Hence, the integral \[ q_0 (x) \int_e^N \frac{\sin (\xi x)}{\xi}\,{\text d}\xi \] admits an estimate \[ \left\vert q_0 (x) \int_e^N \frac{\sin (\xi x)}{\xi}\,{\text d}\xi \right\vert \le \left\vert q_0 (x) \right\vert . \] Since q₀ ∈ 𝔏&sip1;([0,∞)), Riemann--Lebesgue lemma allows us to take the limit under the sign of integral \[ \lim_{N\to\infty} \int_e^N \frac{\sin (\xi x)}{\xi}\,{\text d}\xi = \int_e^{\infty} \frac{\sin (\xi x)}{\xi}\,{\text d}\xi < \infty . \] We got a contradiction because the integral in the left hand-side diverges. Therefore, our assumption that there exists q(x) which Fourier transform in Q, is wrong! ■
Minkowski’s Inequality: If f and g are 𝔏∞ functions, then f + g is 𝔏∞, and
Hölder's Inequality: If f is an 𝔏¹ function and g is an 𝔏∞ function, then their product f g is
Lebesgue integrable and
This example also shows that the Fourier operator ℱ : 𝔏¹(ℝ) → 𝔏∞(ℝ) is continuous. The inequality
\[
\| \hat{f}\|_{\infty} \le \| f \|_1
\]
says exactly that the operator norm of ℱ (viewed as a map from 𝔏¹(ℝ) to 𝔏∞(ℝ)) is at most 1. In other words, if fₙ → f in 𝔏¹, then
\[
\| \hat{f}_n - \hat{f} \|_{\infty} \le \| f_n - f\|_1 \to 0 ,
\]
so Fourier transforms ℱ[fₙ] converges to ℱ[f] in 𝔏∞.
■
We start from the Fourier transform of f.
Since f = F′, we compute
\[
\hat{f}(\xi ) = \int_{\mathbb{R}} F′(x)\,e{−\mathbf{j}x\xi} \,{\text d}x.
\]
Use integration by parts with
\[
\begin{split}
u&= e^{−\mathbf{j}x\xi}, \quad \Longrightarrow \quad {\text d}u = −\mathbf{j}\xi\,e^{−\mathbf{j}x\xi}\, {\text d}x, \\
{\text d}v &= F′(x)\,{\text d}x, \quad \Longrightarrow \quad v=F(x).
\end{split}
\]
Then
\[
\hat{f}(\xi ) = \int_{\mathbb{R}} F(x)\,e^{−\mathbf{j}x\xi}\,\mathbf{j}\xi\,{\text d}x .
\]
Vanishing of the boundary term.
We claim
\[
\lim_{x\to\pm\infty} \,F(x)\, e^{−\mathbf{j}x\xi} = 0.
\]
This follows because:
We make a conclusion:
So we obtain
\[
\hat{f}(\xi ) = \mathbf{j}\xi\,\hat{F}(\xi ).
\]
For ξ ≠ 0, rearranging gives
\[
\hat{F}(\xi ) = \mathbf{j}\xi\,\hat{f}(\xi ).
\]
Final remark
The identity holds for all ξ ≠ 0. At ξ = 0, the expression jξ1 is undefined, and the relation must be interpreted separately (typically in a limiting or distributional sense).
This completes the proof.
This example explicitly verifies the theorem’s formula on a nontrivial pair , and also illustrates the subtlety at ξ = 0, where the identity holds in the sense of continuous extension.
■
Let us consider the Fourier transformation of the characteristic function
\begin{align*}
\Phi (a,b,\xi ) &= \int_{-\infty}^{\infty} \chi_{[a,b]} (x)\, e^{-{\bf j}x\xi} {\text d} x = \int_a^b e^{-{\bf j}x\xi} {\text d} x
\\
&= \frac{e^{-{\bf j}b\xi} - e^{-{\bf j}a\xi}}{-{\bf j}\xi}
\end{align*}
This formula shows that the Fourier transform of a characteristic function is continuous on all real axis and
\[
| \Phi (a,b,\xi )| = \frac{|e^{-{\bf j}b\xi} - e^{-{\bf j}a\xi} |}{|-{\bf j}\xi|} \le \frac{2}{|\xi |} .
\]
This gives
\[
\lim_{\xi\to\infty} | \Phi (a,b,\xi )| = 0 .
\]
Any step function is a linear combination of characteristic functions:
\[
h(x) = \sum_{k=1}^n \alpha_k \chi_{[a_k , b_k ]} (x) .
\]
Then its Fourier transform is
\begin{align*}
\hat{h} &= ℱ[h] = ℱ\left\{ \sum_{k=1}^n \alpha_k \chi_{[a_k , b_k ]} (x) \right\}
\\
&= \sum_{k=1}^n \alpha_k ℱ\left[ \chi_{[a_k , b_k ]} \right]
\\
&= \sum_{k=1}^n \alpha_k \frac{e^{-{\bf j}b_k \xi} - e^{-{\bf j}a_k \xi}}{-{\bf j}\xi} ,
\end{align*}
which can be estimated
\[
\left\vert \hat{h} \right\vert \le \frac{2}{|\xi |}\sum_{k=1}^n \left\vert \alpha_k \right\vert .
\]
Therefore, the Riemann--Lebesgue lemma is valid for step functions. It is known that the class of step functions is dense in 𝔏¹, which means that any Lebesgue integrable function f can be approximated with step functions: \( \displaystyle \quad h(x) = \lim_{n\to\infty} h_n (x) , \quad \) for some sequence of step functions (hₙ), namely,
\[
\lim_{n\to\infty} \| f - h_n \| = 0 .
\]
For any ε > 0, there exists positive integer N such that for all n ≥ N,
\[
\| f - h_n \| < \varepsilon .
\]
We estimate the Fourier transform of this difference
\begin{align*}
\| \hat{f} - \hat{h}_n \| &= \left\vert \int_{\mathbb{R}} \left[ f(x) - h_n (x) \right] e^{-{\bf j} x\xi} {\text d} x \right\vert
\\
& \le \int_{\mathbb{R}} \left\vert f(x) - h_n (x) \right\vert {\text d} x = \| f - h_n \|_1 < \varepsilon , \quad n \ge N .
\end{align*}
This inequality shows that ℱ[f] is a uniform limit with respect to ξ ∈ (−∞, ∞) of a sequence of continuous functions that vanish at infinity. Hence, ℱ[f] posses all properties expressed in Riemann--Lebesgue lemma. Indeed, for any ε > 0, there exists N > 0 such that for all n ≥ N, we have
\[
\left\vert \hat{f} - \hat{h}_n \right\vert < \varepsilon \qquad \forall n \ge N .
\]
Then
\[
0 \le \left\vert \hat{f} \right\vert < \left\vert \hat{h}_n \right\vert + \varepsilon
\]
For this ε > 0, find positive number M such that from inequality |ξ| > M follows inequality \( \displaystyle \quad \left\vert \hat{h}_n \right\vert < \varepsilon . \ \) In this case, the relation \( \displaystyle \quad 0 \le \left\vert \hat{f} \right\vert < \left\vert \hat{h}_n \right\vert + \varepsilon \ \) becomes
\[
0 \le \left\vert \hat{f} \right\vert < 2\varepsilon ,
\]
if |ξ| > M. This is equivalent to
\[
\lim_{|\xi | \to\infty} \left\vert \hat{f} (\xi ) \right\vert = 0 .
\]
hence F(x) → 0 as ∣x∣ → ∞ (a standard result: absolutely continuous functions with integrable derivative and integrable value must vanish at infinity).
Thus, the boundary term is zero:
\[
\left[ F(x)\,e^{−\mathbf{j}x\xi} \right]_{x=-\infty}^{x=\infty} = 0.
\]
Separating real and imaginary parts, we conclude that for any absolutely integrable function f,
Compute its Fourier transform: \[ \hat{f}(\xi ) = \int_0^1 e^{−\mathbf{j}x\xi} \,{\text d}x . \] This integral is elementary if you have an access to a computer algebra system: \[ \hat{f}(\xi ) = \left[ \left( −\mathbf{j}\xi \right)^{-1} e^{−\mathbf{j}x\xi} \right]_{x=0}^{x=1} = \frac{\mathbf{j} \left( \cos\xi -1 \right) + \sin\xi }{\xi} , \quad \xi \ne 0. \]
Now we analyze the asymptotic behavior. We estimate the magnitude: \[ \left\vert \hat{f}(\xi ) \right\vert = \left\vert \frac{\mathbf{j} \left( \cos\xi -1 \right) + \sin\xi }{\xi} \right) \le \frac{3}{|\xi |} . \] Hence, \[ ∣\hat{f}(\xi ) \mapsto 0 \quad \mbox{as } ∣\xi ∣ \to \infty . \]
This example clearly demonstrates the Riemann–Lebesgue phenomenon:
The function f is localized in space (compact support). Its Fourier transform is spread out but decays to zero. The decay here is of order O(1/∣ξ∣), reflecting the jump discontinuities of f.Remark: Smoother functions exhibit faster decay. For instance, if f is continuously differentiable with f′ ∈ 𝔏¹, then \[ \hat{f}(\xi ) = O\left( ∣\xi ∣^{-1} \right) , \] and higher smoothness yields even faster decay. This example provides a concrete and computable instance of the Riemann–Lebesgue Lemma in action. ■
- Carslaw, H.S., Introduction to the theory of Fourier's series and integrals, Third edition, Dover Books on Advanced Mathematics, 1950.
- Titchmarsh, E.C., Introduction to the theory of Fourier integrals, Oxford at the Clarendon Press, 1937.
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