# Tag Info

5

It's not quite as simple as that. The union $$N = \bigcup_{j,k,m \in \Bbb{N}} N_{j,k,m}$$ is a countable union, so properties of measures hold. If we take the most straightforward implementation of your replacement scheme, we need $$N = \bigcup_{\substack{j,k\in \Bbb{N} \\ 0 < \varepsilon < \varepsilon_0}} N_{j,k,\varepsilon}$$ which is no ...

4

Start with an orthonormal set $\{ f_n \}_{n=1}^{\infty}$ in $L^2[1/2,1]$, and extend the functions to be $0$ on $[0,1/2]$ in order to obtain an orthonormal set $\{ \tilde{f}_n \}_{n=1}^{\infty}$ in $L^2[0,1]$. Then, for $n\ne m$, \begin{align} \|T\tilde{f}_n-T\tilde{f_m}\|^2 & =\|x\tilde{f}_{n}-x\tilde{f}_m\|^2 \\ & ...

4

Lemma: Assume $p>1.$ Let $c\in \mathbb R.$ Then there exists a sequence $f_n\in L^1\cap L^p$ such that $\int f_n = c$ for all $n,$ and $\|f_n\|_p \to 0.$ Let's assume the lemma. Let $f\in L^1\cap L^p$ and put $c= - \int f.$ By the lemma, there are $f_n \in L^1\cap L^p$ such that $\int(f+f_n) = 0$ for all $n,$ with $\|f_n\|_p \to 0.$ Then $f+f_n \to f$ in ...

4

Obviously, $||f||_{1,\infty} = \sup_{\alpha>0} \alpha \lambda_f(\alpha)$. For $n=1$ and a monotonic $f$ this is the area of the largest rectangle under the graph of $f$. The idea is thus the following. Take $f$ such that this rectangle has a large horizontal side. We add a small $g$, which bumps $f$ at the point where the rectangle touches the graph, ...

4

You are definitely on the right track, indeed Holder inequality is all you need to apply: $$\|f\|_r^r = \int_E |f|^r = \int_E |f|^r\cdot 1 \le \Big(\int_E |f|^{r\frac pr}\Big)^{\frac rp}\Big(\int_E 1\Big)^{1 - \frac rp} = \|f\|_p^{\frac rp}m(E)^{1 - \frac rp}.$$ Raising both sides to the power $\frac{1}{r}$ we obtain the desired inequality. Notice that ...

3

No, if you allow $||x_\infty||_1=\infty$: let $$x_n(j)= \begin{cases} 1/j,&(1\le j\le n), \\0,&(j>n).\end{cases}$$ Yes if $||x_\infty||_1<\infty$. Let $\epsilon>0$. Choose $A$ so $$\sum_{j=A+1}^\infty|x_\infty(j)|<\epsilon.$$Since convergence in $\ell^2$ implies pointwise convergence there exists $N$ so that $$\left|\vert \vert ... 3 S is not closed. Consider$$f_n(x)=\min\left(\left\lvert n\left(x-\frac12\right)\right\rvert,1\right)$$f_n\in C[0,1] and$$\forall n\ge 2,\quad\int_0^1 \lvert 1-f_n(x)\rvert\,dx =\frac1n$$So f_n\to 1 in L^1 norm, but 1\notin S. It is actually dense: you can show that, for all g\in C[0,1], f_n\cdot g\to g in L^1 norm (recall that \lVert ... 3 No, certainly not. Just consider f=2\cdot 1_{[0,1]}+1_{[2,10]} and g=1_{[0,1]}+1000\cdot 1_{[2,10]}. 2 Define g_M := \max(\min(g,M),-M) ∈ L^1 ∩ L^\infty and apply your argument modified so that s_k → \operatorname{sgn} g_M in L^1 to find that ‖g_M‖_1 = 0. On the other hand, |g_M| \uparrow |g| ∈ L^1, so MCT says ∫|g_M| → ∫ |g|, which means ∫|g| = 0. (Actually since g_M^2 \uparrow g^2 as well you don't need to modify your s_k but I'll just ... 2 Yes, it is correct. The operators T_N are of finite rank (hence compact) and converge to T. Thus, T is compact as well. 2 By the monotone convergence theorem, you always will have$$\int f^p = \lim_{k \rightarrow \infty} \int f_k^p$$Taking pth roots (and using continuity of the function x \rightarrow x^{1 \over p}), one therefore has$$||f||_p = \lim_{k \rightarrow \infty} ||f_k||_p \tag 1$$Since the f_k increase to f, the limit in (1) is an increasing limit. So one ... 2 Consider the non-negative function$$F(x):= \sum_{n\geq1}\int_{n}^{n+n^{-\alpha}}|f(x+y)| \, dy$$and notice that if we manage to prove that F\in L^{1}(\mathbb{R}) the problem is settled. Now to lets rewrite F by making the change of variable y=t+n. We will simply get$$F(x)=\sum_{n\geq1}\int_{0}^{1}\chi_{[0,n^{-\alpha}]}(t)|f(x+t+n)| \, dt$$Now ... 2 If you're not going to L^4 (since there is no good reason to...), in the last inequality you get$$ t^{-6/5}\|x^{-2/5}f\chi_{[0,t]}\|_5\|x^{2/5}\chi_{[0,t]}\|_{5/4}. $$A simple integration gives \|x^{2/5}\chi_{[0,t]}\|_{5/4}=t^{6/5}(2/3)^{4/5}, so you get$$ (2/3)^{4/5}\|x^{-2/5}f\chi_{[0,t]}\|_5, $$which goes to 0 indeed, by DCT. 2 For the first one, it is indeed just the Riemann-Lebesgue Lemma: If f\in L^q[-\pi,\pi], then f\in L^1[-\pi,\pi]. Hence$$\int_{-\pi}^\pi f(x) \sin (kx)\, dx \to 0$$by RL. So  \sin (kx) \to 0 weakly in L^p as desired. For the second one, we don't need to evaluate the integral exactly. Instead we can just note$$\int_0^\pi |\sin (kx)|^p\, dx = ...

2

Consider the Rademacher functions $r_n(x) = \operatorname{sign}\sin 2^n x$, restricted to $[0, 2\pi]$ interval. Translating by $\pi/2^n$ changes $r_n$ almost into $-r_n$, except for two pieces of size $\pi/2^n$ on the boundary. Hence, the $L^2$ norm of $S(\pi/2^n)r_n-r_n$ is almost twice the norm of $r_n$. This extends to general translation amounts $t$ ...

2

Let's break things down into steps: $L^p$ is a vector space. (Follows from their definition.) They are normed vector spaces: The $L^p$ norm, by definition, is a finite, nonnegative real number for given $f \in L^p$. 2.1 $\|f\|_p=0$ iff $f=0$ in $L^p$. This follows from the fact that if $f\neq 0$ on a set of positive measure, then $\int |f|^p >0.$ 2.2 ...

2

Yes, you can use the quoted theorem. We can also argue only with the dominated convergence theorem. Indeed, for a fixed $R$, we have $$\left|\int f_i^2-\int f^2\right|\leqslant \left|\int f_i^2\mathbf 1\{|f_i|\leqslant R\}-\int f^2\mathbf 1\{|f|\leqslant R\}\right|\\ +\left|\int f_i^2\mathbf 1\{|f_i|\gt R\}-\int f^2\mathbf 1\{|f|\gt R\}\right|\\ \leqslant ... 2 Following Daniel Fischer comments I'm trying to post an answer: Let G = (a,b) We have that$$\lVert u\rVert_{L^\infty(G)}^2 = \lVert u^2\rVert_{L^\infty(G)} \le \int_a^b 2 |u(t)u'(t)|dt \le 2 \lVert u\rVert_{L^2(G)}\lVert u'\rVert_{L^2(G)}$$where the first inequality is justified by the fact that u^2 is absolutely continuos and u(a) = 0 so that we ... 2 You are definitely on the right track: I would consider something like$$ f = \begin{cases} \frac{1}{x^a(\log x)^b} & \text{if}\ x \in (0,0.5) \\ 1 &\ \text{if}\ x \in [0.5,1) \end{cases} $$so that you don't need to worry about integrability at 1. Now you only need to play around with a and b. A choice that works is the following: 2 Now that I've actually read it carefully, let me try again. He throws away those A_k in order to be able to say that f is bounded on E^c. Your proof appears correct as well. My guess is that the main reason for using those A_k was to shorten the proof, because this way he didn't have to use any epsilons at all. Your f may not be bounded on your ... 1 I'm not exactly sure what level of rigour you're aiming for; the problem may at least in part be that you're expecting more rigour than physicists usually care for; but here are two (possibly related) points at which I don't follow your objections: You write that the Euler-Lagrange equation is f''(x)=U(x)f(x) – that's missing the term Ef(x), which ... 1 If f_n(x) =|x-1/2|^{1/n}, then each f_n \in S, but f_n \to 1 in the L^1 norm. 1 The phrasing "by invoking transfinite induction" is indeed rather cryptic here. In the case p=\infty, there is no obvious way to use transfinite induction to modify the given argument to work for non-\sigma-finite \Omega. I would guess that the book is instead attempting to allude to the Hahn-Banach theorem, which implies that the statement of the ... 1 Let f\in L^1\cap L^2 and f_t(x)=t^n\,f(t\,x). Then f_t\in L^1\cap L^2 and$$ \|f_t\|_1=\|f\|_1,\quad \|f_t\|_2=t^{n/2}\,\|f\|_2. $$As t\to0, f_t converges to 0 in L^2, but not in L^1. The example can be changed to cover the case L^p\cap L^q. 1 For part a): if g\in C^1[-\pi,\pi],$$ \int_{-\pi}^\pi g(t)\sin kt\,dt=\left.-\frac1k\,g(t)\,\cos kt\right|_{-\pi}^\pi+\frac1k\int_{-\pi}^\pi g'(t)\,\cos kt =\frac1k\int_{-\pi}^\pi g'(t)\,\cos kt $$As g' is continuous in the interval, it is bounded. The integral on the right is thus bounded, and so the whole thing goes to zero as k\to\infty. If now ... 1 The reason it's possible for A^c to be dense is that Sx=\sum x_n is not a bounded linear functional on \ell^2. So you can have sequences with zero sum converging in \ell^2 to a sequence with non-zero sum; no contradiction because S is not continuous. In fact, since \sum x_n need not converge for x\in\ell^2 it seems to me that the problem is ... 1 Concerning 3., do not estimate the integrals \int f_j(x) \phi(x+n)dx in your way but approximate f_j by functions with compact support (just multiply with an indicator function). If f_j has compact support the integral is 0 for n large enough. Concerning 2. the limit of a convergent subsequence would be necessarily 0 because of 3. But the norms ... 1 You can show your regulary if g \in W^{1,p}(\Omega), p > d \ge 2, by using a trick by Stampacchia. First, we note$$a(u - g,v) + \int_\Omega f(u) v \, \mathrm{d} x = a(-g,v).$$Now, let K > 0 and set$$(u - g)_K = \begin{cases} u - g & \text{if } |u - g| \le K, \\ \operatorname{sign}(u-g) \, K & \text{else}.\end{cases}$$Then, (u-g)_K ... 1 Use the useful inequality$$\left\|f\cdot g\right\|_p\leq\left\|f\right\|_p\cdot\left\|g\right\|_{\infty}

1

Take $f \in L^p\setminus L^q$ And define $E_n=\{x \in X: |f(x)|\geq n\}$ To show that $E_n$ has positive measure use Minkowski and the fact that $||f \chi_{E^c_n}||_q\leq ||f \chi_{E^c_n}||_p^{p/q} || f \chi_{E^c_n}||_{\infty}^{1-p/q}$ Showing that $\int|f|^{1/q}*\chi_{E_n} d\mu =\infty$, and consequently $E_n$ Has measure diferent then 0, ie. Strictly ...

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