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6

No, this doesn't follow. Fix a smooth bump function $\varphi$ supported in $[-1,1]$ and let $$ f(x) = \sum_{n=1}^\infty \phi(2^n x-n) $$ The $n$th term is supported in $[n-2^{-n}, n+2^{-n}]$; these supports are disjoint. The series converges in every $L^p$ space. The function $f$ does not tend to zero at infinity, through. On the Fourier side, $$ \hat ...


4

First, as pointed out in the comments above, the right hand side needs a factor of $(2\pi)^{-1}$. With that in mind, here we go. Let's pick our favorite $2\pi$ periodic functions. Namely, for $j=0,1,2,\dots$ let $f_j(x)=e^{ijx}$ (where $i$ is the imaginary constant). Now let's see if the equality holds for these functions. For $j=0$, it is easy to see that ...


3

From Parseval's Theorem, we have $$\frac{2}{\pi}\int_0^{\pi}f(x)^2dx=\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)^2$$ Since the series converges, then $f$ is in $\mathscr{L}^2$ We can show this equality as follows $$\begin{align} \int_0^{\pi}f^2(x) &=\int_0^{\pi}f(x)\sum_{k=1}^{\infty}\frac{\sin kx}{k+1}dx \tag 1\\\\ ...


2

It is just a change of variable: $$ \int_{-\infty}^\infty h(-\tau)\,e^{i\tau t}\,d\tau=\int_{-\infty}^\infty h(\tau)\,e^{-i\tau t}\,d\tau=\int_{-\infty}^\infty h(\tau)\,\overline{e^{i\tau t}}\,d\tau=\overline{\int_{-\infty}^\infty h(\tau)\,e^{i\tau t}\,d\tau} $$ because $h$ is real valued.


2

Just a sketch of a possible approach, quite close to one of the usual proofs of the Fejer-Jackson inequality. By periodicity and symmetry, we just need to prove the inequality over $(0,\pi)$. We have: $$\sum_{n=1}^{N}\frac{\sin(nx)}{n}=\int_{0}^{x}\sum_{n=1}^{N}\cos(nt)\,dt=-\frac{x}{2}+\int_{0}^{x/2}\frac{\sin((2N+1)u)}{\sin(u)}\,du \tag{1}$$ and we may ...


2

The change $$ u(t,x)=e^{ax+bt}\,v(t,x) $$ for appropriate choice of $a,b\in\mathbb{R}$ will transform the equation into $$ v_t=v_{xx},\quad v(0,x)=e^{-ax}\,g(x). $$


2

It helps to recall Which functions are tempered distributions? — exactly the distributional derivatives of continuous functions of polynomial growth. The full statement of this theorem isn't needed here, but it gives an idea where to look for weird tempered distributions that are represented by a locally integrable function: take the derivative of a ...


2

No. The function $$ f(x)=\begin{cases}0 & |x|>1\\ (1-\log(1-x^2))^{-1} & |x|\le1\end{cases} $$ is an explicit example. For more comments and a proof (due to Terry Tao) see this post.


2

No, there exist continuous functions on $S^1$ whose Fourier coefficients are not in $l^{\beta}$ for every $\beta <2$ ( while certainly being in $l^2$). The first example seems to be given by T. Carleman in 1916. This should be discusses in some books on classical Fourier series. I learned about it from this paper of Hausdorff.


1

The Riemann-Lebesgue Lemma states that if $f$ is $\mathscr{L}^1$ integrable, then $$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=0$$. Obviously, for real-valued $f$, we have $$\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)e^{i\omega t}dt=\lim_{\omega \to \infty}\int_{-\infty}^{\infty}f(t)\cos(\omega t)dt+i\lim_{\omega \to ...


1

There may be a misunderstanding here; I don't see how one would use Jensen's inequality to control $L^p$ norm by $L^1$ norm. Anyway, the proof of $$ \lim_{n\to\infty}\|f-h_n*f\|_p =0 $$ is essentially the same for $1<p<\infty$ as it is for $p=1$. For continuous functions $g$ we have $h_n*g\to g$ uniformly: just split the integral in ...


1

If you consider a minimal state space representation of $G$ $$\dot{x}=Ax+Bu\\y=Cx+Du$$ then $A$ must be a stable (Hurwitz) matrix for stable $G$. If $u\in\mathcal{L}_2$ and $A$ is stable then $x\in\mathcal{L}_2$ Proof: If $A$ is stable there exists $P=P^T>0$ such that $$PA+A^TP=-I$$ Define now the nonnegative function $V=x^TPx$. Then ...


1

Let us start with a general claim: Let $(g_n),\quad n=1,2,\ldots$, be a sequence of functions which converges uniformly to the function $g$. Let $h$ be a square-integrable function. Then the sequence $(g_nh)$ converges by integral to $gh$. To prove the claim, we may assume $g=0$. By the Cauchy-Schwarz inequality we have $$\left(\int g_nhdx\right)^2\leq\int ...



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