Tag Info

Hot answers tagged

3

To derive Nash's inequality from $$\|f\|_2^2\leq 2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2\tag{1}$$ you should choose the value of $\lambda$ that minimizes the right hand side. To this end, consider the function $$g(\lambda)=2\lambda \|f\|_1^2+\frac{1}{4\pi^2\lambda^2}\|f'\|_2^2$$ of a real variable $\lambda$. Then ...


3

Let $\mathbf A = \left\{a_{ij}\right\}_{i,j=0}^n \in \mathbb C^{(n+1) \times (n+1)}$ be defined as $$ a_{ij} = c_{i+j \operatorname{mod} (n+1)}, $$ so $$ \mathbf A = \begin{pmatrix} c_0 & c_1 & \dots & c_n\\ c_1 & c_2 & \dots & c_0\\ \vdots & \vdots & \ddots & \vdots\\ c_n & c_0 & \dots & c_{n-1} \end{pmatrix}. ...


3

This is called an anticirculant matrix, which is a special case of Hankel matrix. The eigenvalue/eigenvector formula for circulant matrix does not apply.


2

Yes, there exist such functions. Let for instance $$ f(x)=\begin{cases}\sin x & 0\le x\le\pi,\\g(x) & -\pi\le x<0,\end{cases} $$ where $g\colon[-\pi,0]\to\mathbb{R}$ is to be chosen in such a way thet $f$ is continuous and piecewise $C^1$ on $[-\pi,\pi]$ and $$ \int_{-\pi}^0g(x)\cos(k\,x)\,dx=-\int_0^\pi \sin x\cos(k\,x)\,dx,\quad k=0,1,2 $$ and ...


2

Well, we saw the other day that the observation that $1/N=N^2/N$ leads to a counterexample. Nothing like making a blithering idiot of oneself for motivation.... We will take $2\pi$ and perhaps other absolute constants to have the value $1$. The letters $I$ and $J$ will always denote intervals; as usual, $|I|$ is the length of $I$. Lightbulb: If $I$ and $J$ ...


1

If $f$ and $f'$ are in $L^{2}$, then $ff'$ is absolutely integrable because $$ 2|ff'| \le |f|^{2}+|f'|^{2}. $$ Therefore, in this case, the following has finite limits as $x\rightarrow\pm\infty$: $$ f^{2}(x)-f^{2}(0)=2\int_{0}^{x}f(t)f'(t)dt. $$ In this case, the limits $\lim_{x\rightarrow\pm\infty}f^{2}(x)=L_{\pm}$ must be $0$ because, ...


1

I am not familiar with your context, but I may be able to help out understand the definition and implications on $\lim_{x \rightarrow \pm \infty} f(x)$. A function is square integrable on $(-\infty, \infty)$, hence the name, if $$ \int_{-\infty}^\infty f(x)^{\color{red}{2}} \mathrm{d}x \lt \infty $$ (… and the above makes actually sense, ie. $f$ is ...


1

This is not a complete answer but I want show that there is nothing mysterious going on here. We want to prove that: $$\text{Fourier Transform of } \Lambda(1)...\Lambda(k) \sim \sum\limits_{n=1}^{n=\infty} \frac{1}{n} \zeta(s)\sum\limits_{d|n}\frac{\mu(d)}{d^{(s-1)}}$$ The Dirichlet inverse of the Euler totient function is $$a(n)=\sum\limits_{d|n} ...


1

Yes, you've used it correctly. It's true that $|\hat{f}| \le \|f\|_{L^\infty}$ only almost everywhere, but that is sufficient for your conclusion. To see why, let $A = \{|\hat{f}| \le \|f\|_{L^\infty}\}$, and break your integral into two integrals, over the sets $A$ and $A^c$. Then note that $A^c$ has measure zero. Intuitively, the Lebesgue integral ...


1

To get a concrete simple example, think of a function $f$ expanded in $[0,\pi]$ in a sine-only Fourier series (it is enough to think of $f$ extended to $[-\pi,0]$ as an odd function). Coefficients $b_k$ of this Fourier series are given by $b_k=\int_0^\pi f(x)\sin(kx)\,dx$ (apart a normalization factor). It is then enough to choose $f$ so that $b_1=b_2=0$. ...


1

A bounded periodic integrable function F will certainly "have" a Fourier series, but the sum of the series can fail to be equal to F at some points, even if F is continuous.


1

$$\kappa(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s) = \int_0^\infty x^{s-1} \sum_{n=1}^\infty e^{- \pi n^2 x^2} dx$$ (it is what Mike calls the the Gamma-completed Riemann zeta function. it is useful because $\kappa(s) = \kappa(1-s)$) you know that $\displaystyle f(x) = \sum_{n=1}^\infty e^{- \pi n^2 x} = \sum_{m=1}^\infty \frac{(-1)^{d_p(m)}}{e^{\pi mx}-1}$ ...


1

This can be done as follows, given $f$ define $$I(t) = f(t)\cos(\Omega t) \,\,\,\,\,\,\,\, Q(t) = -f(t)\sin(\Omega t)$$ Then $$I(t)\cos(\Omega t) - Q(t)\sin(\Omega t) = f(t)\cos^2(\Omega t) + f(t)\sin^2(\Omega t) = f(t)$$


1

No, thay cannot be naturally generalised to exponents $p>2$. A random variable $X$ is $p$-stable if whenever we have a sequence $(X_n)_{n=1}^\infty$ of independent copies of $X$ then for any finite sequence of scalars $(\alpha_n)_{n=1}^N$ the random variables $$\sum_{n=1}^N \alpha_n X_n\quad\text{ and }\quad\Big(\sum_{n=1}^N |\alpha_n|^p\Big)^{1/p}X$$ ...


1

Tempered distributions have Fourier transforms. If $P(D)\,F=0$, take the Fourier transform to get $P(\xi)\,\hat F(\xi)=0$ for all $\xi\in\mathbb{R}$. Since $P(\xi)\ne0$ if $\xi\ne0$ the support of $\hat F$ is $\{0\}$. Can you take it fromhere?


1

There is some ambiguity in this problem, just as you pointed out in your question. If you start with the Fourier transform of your equation in $x$, which is given by $$ \hat{u}(s,y) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-isx}u(x,y)dx, $$ then the equation in $y$ is informally \begin{align} \frac{\partial^{2}}{\partial ...



Only top voted, non community-wiki answers of a minimum length are eligible