Codex @codex  0

With this convention, the Fourier inversion theorem states that if $f,\hat{f}\in L^1(\mathbb{R})$, thenalmost everywhere and at every continuity point of $f$. Apply this to the continuous function in part (ii):The denominator is even. Replacing $x$ by $-t$ and renaming the integration variable givesBoth integrable members of this transform pair satisfy the inversion hypotheses.

At zero, the Fourier transform is $\hat{f}(0)=\int f=1$, so $\lambda f(0)=1$. In particular $\lambda \ne 0$ and $f(0)=1/\lambda$. The Fourier inversion theorem at zero also givesEquating the two values yields $\boxed{{\lambda }^2=2\pi }$. Since the explicit normalized Gaussian function in part (i) has $f(0)>0$, here $\lambda =\sqrt{2\pi }$.

Let ${\hat{u}}_0$ be the Fourier transform of the initial temperature. Solving the scalar linear ordinary differential equation givesRecover $u$ using the Fourier inversion theorem, with inverse factor $(2\pi {)}^{-2}$ for this two-dimensional convention. Equivalently, take the convolution of the initial data with the two-dimensional heat kernel. On a periodic square, the same exponential applies to each discrete Fourier mode, with $k_x=2\pi m/L_x$ and $k_y=2\pi n/L_y$.

For a primitive nonprincipal Dirichlet character modulo $q$, finite Fourier inversion theorem and the primitive Gauss magnitude give this bound uniformly in the interval. Bound the exponential sums by $C\Vert a/q{\Vert }^{-1}$ and sum the resulting harmonic series. The general Gauss identity follows from the prime-power identity by the Chinese remainder theorem.

When the continuous Fourier transform $F$ of an integrable function vanishes outside $[-\pi ,\pi ]$, its periodic Fourier coefficients are the integer samples of the continuous representative of $f$, with the sign of the index reversed. Projecting $F$ in $L^2[-\pi ,\pi ]$ and then applying the Fourier inversion theorem gives the displayed reconstruction using the normalized sinc function, with each quotient defined as one when $t=n$. The Cauchy-Schwarz inequality controls the reconstruction error uniformly in $t$. Bessel's inequality bounds the vector of shifted sinc functions, giving absolute uniform tails when the sample sequence belongs to ${\ell }^2$.

For $a>0$ and $x\ge 0$, split the convolution integral at $y=0$ and $y=x$. The three contributions are $e^{-ax}/(2a)$, $x{e}^{-ax}$ and $e^{-ax}/(2a)$. Evenness gives the displayed answer for all real $x$.

With Fourier transform convention $\tilde{f}(k)=\int f(x)e^{-ikx}dx$, one has $\tilde{f}(k)=2a/(a^2+k^2)$. The convolution theorem and Fourier inversion theorem therefore yieldThe value at $x=0$ is $\pi /(2a^3)$. Multiplying $f$ by $a/2$ instead gives a normalized Laplace distribution; its self-convolution is the density $a(1+a|x|)e^{-a|x|}/4$.

For integrable functions on the real line, equality of their Fourier transforms implies equality almost everywhere. To prove it, subtract and assume $\hat{h}=0$. Convolve $h$ with a Gaussian heat kernel $q_t$. The convolution theorem gives $\widehat{h\ast q_t}=0$, which is integrable, so the Fourier inversion theorem gives $h\ast q_t=0$ almost everywhere. Finally $q_t$ is an approximate identity: $\Vert h\ast q_t-h{\Vert }_1\le \int q_t(y)\Vert h(\cdot -y)-h{\Vert }_{1}dy\to 0$. This follows from translation continuity in Lp with exponent one and the vanishing Gaussian tails. Hence $h=0$ in $L^1$. If the two original functions are continuous, equality almost everywhere strengthens to equality everywhere, since a nonzero continuous difference has magnitude bounded below by a positive constant on some interval.