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4

Using Plancherel theorem for $y\not=0$ $$\int_{\mathbb R} \left |\frac{\partial u(x,y)}{\partial x}\right |^2 dx=\int_{\mathbb R} \left |\widehat{\frac{\partial u(\cdot,y)}{\partial x}}(\xi)\right |^2 d\xi=\int_{\mathbb R} |\xi|^2 \left |\widehat{u(\cdot,y)}(\xi)\right |^2 d\xi=2\pi\int_{\mathbb R} |\xi|^2 |\widehat{P_y} |^2 |\hat f|^2d\xi$$ From the ...

4

For $f,g\in L^2$, the convolution $f*g$ belongs to $C_0(\mathbb R)$. Taking the Fourier transform of a $C_0$ function is problematic: such transforms are generally not functions, but merely distributions (as Jose27 points out). It is better to work from right to left. The product $\hat f \hat g$ is in $L^1$. Apply the inverse Fourier transform to it ...

3

As stated, the proposition doesn't hold. Consider the $2\pi$-periodic extension $f$ of $\chi_{[-\pi/2,\pi/2]}$. We have \begin{align} \hat{f}(n) &= \frac{1}{2\pi}\int_{-\pi}^\pi f(t) e^{-int}\,dt\\ &= \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} e^{-int}\,dt\\ &= \frac{i}{2\pi n}\left(e^{-in\pi/2} - e^{in\pi/2}\right)\\ &= \frac{i((-i)^n - ... 3 The idea is to evaluate the rest of the integrand at the point where the argument of \delta vanishes (this only works if the argument of \delta is of the form x-c, by the way). The argument vanishes where x-4 = 0, i.e., at x=4, so evaluate the rest of the integrand at 4. There is a more general formula for handling the case where the argument of ... 3 First, we recall the Mellin transform of a function f F(s) =\int_{0}^{\infty} x^{s-1}f(x) dx .$$Now, making the change of variables u=e^{x} in the original integral gives$$I = \int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{-iwx}}{{\rm e}^{x}}}{ \left( 1+{{\rm e}^{x}} \right) ^{2}}}{dx}= \int _{0}^{\infty }\!{\frac {{u}^{-iw}}{ \left( ...

2

The answer is yes! In fact $B$ is real analytic. (Not holomorphic, as you normalize.) Thus the Fourier series converges in any sense you want. Rank deficiency is irrelevant, as you can add a suitable multiple of the identity matrix, and pretty much nothing changes. The proof is not trivial. See Kato's Perturbation Theory.

2

In general, for "nice enough" functions (and polynomials are certainly nice enough) $$\int \delta(x) f(x) \,dx = f(0)$$ In some sense, the Dirac-$\delta$ is "infinite" at $x=0$, and $0$ everywhere else, so, if you pair it with other functions in an integral, it does a really good job of singling out the value of the function at $0$. (note that the Dirac ...

2

Observations: $f$ is odd $f$ is continuous on $\mathbb R$, because the series converges uniformly. Together with 1, this implies $f(\pi)=0$. $f$ is a cubic polynomial on $(-\pi,\pi)$, because the Fourier coefficients of $x^k$ involve $1/n^k$ (integration by parts happens $k$ times). Odd cubic polynomials vanishing at $\pi$ are of the form $A(x^3-\pi^2 ... 2 The Fourier transform as a mapping$T\colon S \to S$, where$S$is endowed with the$L^2$-norm, is an isometry (Placherel's theorem). Thus, since$S$is dense in$L^2(\mathbb{R})$, so is its (unique continuous) extension$T_1 \colon L^2(\mathbb{R}) \to L^2(\mathbb{R}). Then $$\lVert g_k - T_1 f\rVert_2 = \lVert T_1 f_k - T_1 f\rVert_2 = \lVert f_k - ... 2 The derivatives of rapidly decreasing (Schwartz class) functions are also rapidly decreasing. Hence we can sum the derivative f' in a like manner,$$G(x) = \sum_{n\in\mathbb{Z}} f'(x-2\pi n).$$The sum converges locally uniformly, hence G is continuous, and$$\begin{align} \int_0^x G(t)\,dt &= \int_0^x \sum_{n\in\mathbb{Z}} f'(t-2\pi n)\,dt\\ ... 2 Well, apparently,F$is not a Schwartz function, but it's in$L^2$. In addition, one can show that this function is (clearly)$L^1_{loc}$and is, in fact, a tempered distribution (if you have problems with proving it, ask in comments). Thus, we can indeed, take a Fourier transform of it in the sense of distributions and in the sense of$L^2$. How we can ... 2 Even though the function is not$L^1$, residue calculus still helps. Put $$f(z) = \frac{z}{1+z^2}.$$ If$\xi < 0$, then$f(z)e^{-ix\xi} \to 0$on a semi-circle in the upper half-plane. Jordans's lemma (details omitted) shows that $$\int_{-\infty}^\infty \frac{xe^{-ix\xi}}{1+x^2}\,dx = 2\pi i \operatorname{Res}\limits_{z=i} f(z)e^{-iz\xi} = i\pi ... 1$$\int\limits_{-\pi}^0e^{-x}\cos(nx)\,\mathrm{d}x=-\int\limits_{-\pi}^0(e^{-x})'\cos(nx)\,\mathrm{d}x =-e^{-0}\cos(n0)+e^{-\pi}\cos(n\pi)-n\int\limits_{-\pi}^0e^{-x}\sin(nx)\,\mathrm{d}x= -1+(-1)^ne^{-\pi}+n\int\limits_{-\pi}^0(e^{-x})'\sin(nx)\,\mathrm{d}x=$$... 1 I think Fourier Transform, in general, can be defined in various ways. Please check the definition your book or you teacher used, with the one that wolfram uses here: http://reference.wolfram.com/mathematica/ref/FourierTransform.html 1 1: As you rightly say,$(x−m)$and$(x−n)∈[0−3]$defines the support of$\phi_m(x)=\phi(x-m)$and$\phi_n(x)=\phi(x-n)$. Which transforms to$x\in[m,m+3]$resp.$x\in[n,n+3]$. The integration goes over the whole of$\mathbb R$, but because of the supports and the multiplication, it reduces to the intersection of those intervalls. If the intersection is ... 1 The transform is$\hat{g}^{(p,q)} = e^{iq\omega}\hat{g}(\omega-p)$, which can be seen by substituting directly into the equation for the fourier transform and substituting$u=t-q$in the integral. This makes the work for$M_\omega$the same as that for$M_t$. I'm not familiar with$\sigma_t$, but if you post the formula I'll work with it. Edit: With PNB's ... 1 Some things here are merely decorations. Modulation (multiplication by$e^{i\omega t}$) does not change$\sigma_t(x)$because it does not change$\rho_x(t)$. Translation (replacing$t$by$t-\tau$) does not change$\sigma_t(x)$either:$\rho_x(t)$shifts by$\tau$, but so does$\mu_t$, so the net effect is zero. (Simply put, central moments of a distribution ... 1 Yes, the Fourier series converges uniformly because the entries of$\mathbf B$are Hölder continuous functions. (I agree with Yiorgos S. Smyrlis that they are actually real analytic, but I'm not going to prove that.) Here is a reference for the implication Hölder continuity$\implies\$ uniform convergence; in the rest of the post I prove that ...

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