New answers tagged

0

First, I prove this where $f$ is continuous. If $f$ is continuous on $[0,1]$ then it is uniformly continuous on $[0,1]$ and is bounded. Let $\varepsilon>0$ be given. Then there is some $\delta_1>0$ such that for all $x_0\in [0,1]$, if $0<|x_0-x|<\delta_1$ then $|f(x_0)-f(x)|<\sqrt[p]{\varepsilon^p/2}$. Since $f$ is continuous on a closed ...


2

If $p<\infty$, I'd use the density of $C_c(0,1)$ in $L^p$, together with the fact that a continuous function with compact support is uniformly continuous.


1

To make Davide Giraudo's comment more clear, here are the steps: $$ \int |f(x)|^p = \int \left((1+x^2)|f(x)|\right)^p\frac{1}{(1+x^2)^p} $$ From here, since $(1+x^2)f(x)$ is bounded (because $f\in \mathcal{S}$), and $\frac{1}{(1+x^2)^p}\leq \frac{1}{1+x^2}\in L^1$ (for $p\geq 1$), one gets $$ \int |f(x)|^p \leq \|(1+x^2)f(x)\|^p_\infty \int ...


1

I think that a simpler proof comes from $$\left|f(x)\right|\leq\frac{K}{1+x^2}\quad \Longrightarrow\quad \int_{\mathbb{R}}\left|f(x)\right|^p\,dx \leq K^p\cdot\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(p-\frac{1}{2}\right)}{\Gamma(p)}.$$


3

No, the sequence $(g_n)$ need not be Cauchy in the $L^2$ topology. Consider $g_1 = \begin{cases} 1, && x \in [0,1] \\ 0, && \text{otherwise}\end{cases}$. Then $g_n = \begin{cases} n, && x \in [0, \frac 1 n] \\ 0, && \text{otherwise}\end{cases}$. One way of showing that $(g_n)$ is Cauchy is showing that $\| g_{n+p} - g_n \| ...


1

Theorem 1: Let $1 \leq p<q \leq \infty$. If $X$ contains sets of arbitrarily small positive measure, then $L^p \setminus L^q$ is nonempty. Theorem 2: Let $1 \leq p<q \leq \infty$. If $X$ contains sets of arbitrarily large finite measure, then $L^q \setminus L^p$ is nonempty. Theorem 3: Let $1 \leq p<q<r\leq \infty$. If $f \in L^p \cap L^r$ ...


3

In fact, given a measure $\mu$, let $$M=\{\mu(E):0<\mu(E)<\infty\}.$$ Then $L^p\subset L^q$ whenever $p\le q$ if and only if $\inf M>0$, and $L^p\subset L^q$ whenever $p\ge q$ if and only if $\sup M<\infty$.


1

Note that in order to show that the function $f \ast g$ is in $L^1$ we just have to integrate its absolute value over the variable, in this case over $x$. For the detailed calculation consider $\int_{\mathbb{R}^n}| (f \ast g)(x)| dx=\int_{\mathbb{R}^n}| \int_{\mathbb{R}^n} f(y) g(x-y) dy| dx \leq \int_{\mathbb{R}^n} \int_{\mathbb{R}^n} |f(y) g(x-y)| dy ...


2

If $\int_E f$ is finite then it necessarily exists. Q: "Is $f$ in $L(E)$ or not if $\displaystyle \int_E f$ is finite but oscillating." A: If $\int_E f$ is oscillating, then $\int_E f$ doesn't exist. And since it doesn't exist, then it can't be finite (contrapositive of "if finite, then exists"). Consider $\int_\Bbb{R} \sin x \, dx$ as a concrete ...


1

The convergence of the series implies that $w(2^k)$ is finite for each integer $k$. Now, observe that if $t\gt 0$, then $2^k\leqslant t$ for some $k\in\mathbb Z$, hence $$w(t)\leqslant w(2^ k)<+\infty.$$


1

Hint: $$|f(x)|^q \le \max(|f(x)|^p,1)$$


1

Let $p>0$. Raising both sides of the inequality to the power $-p$, we find that it is equivalent to $$ \int_0^{2\pi} \left(\frac{|e^{i\theta}+b|}{a+b}\right)^pd\theta \geq \int_0^{2\pi} \left(\frac{|e^{i\theta}+c|}{a+c}\right)^pd\theta. $$ Now to prove this inequality, we show that \begin{equation}\tag{1} \frac{|e^{i\theta}+b|}{a+b}\geq ...


1

$$L^p(E)=\{f: \int_E \vert f \vert^p <\infty \} $$ So $ f\in L^p(E) \rightarrow f^p \in L^1(E)$ Then you can prove your question 1 by contraddiction! ($f^p = \infty$ on a positive measure set then...) About question 2, I thin you're right, there must be an absolute value!


0

Recall that any series $\sum_{n=1}^\infty a_n$ of scalars is convergent if it is absolutely convergent (i.e. if $\sum_{n=1}^\infty |a_n| < \infty$). This fact turns out to be closely related to the fact that the field of scalars ${\Bbb C}$ is complete. This can be seen from the following result: (This is essentially your Proposition 2) Let $(V, \| ...


1

(1) is true, as $f$ is a limit of sequence of functions $f_n$, $f_n = f \cdot \chi_{(-1 + 1/n, 1-1/n)}$, which are all measurable on $(-1, 1)$ by assumption, so $f$ is measurable on $(-1, 1)$ as a limit of functions measurable on $(-1, 1)$. It's also not hard to show that it's measurable directly. For (3), see my comment above. Your argument for (2) looks ...


0

Let $X=[0,1]$. Let $\mu$ be counting measure on $X$, except restricted to the $\sigma$-algebra of sets $E$ such that $E$ or $X\setminus E$ is countable. Define $\Lambda\in(L^1(\mu))^*$ by $$\Lambda f=\sum_{x\in[0,1/2]}f(x).$$ Then $\Lambda$ cannot be represented as integration against any $L^\infty$ function; loosely speaking, it's clear that that function ...


0

Enlightened by discussion in this related question: In Rudin's proof of the completeness of $L^\infty$, I conclude that the proof in "Added2" in OP is correct. More precisely, I rewrite it as follows. Define for each $j,k\in\mathbb{N}$ a null set $N_{j,k}$ such that $$ |f_j(x)-f_k(x)|>\|f_j-f_k\|_\infty\quad\textrm{for all }x\in N_{j,k}. $$ Suppose ...


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 ...


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 ...


0

Let $\varepsilon$ be smaller than $\min\{b-a,1\}$ and $f(x)=g(x):=\mathbf 1_{[a,a+\varepsilon]}(x)$. Then $$\lVert fg\rVert_p=\left(\int_{[a,b]}\mathbf 1_{[a,a+\varepsilon]}(x)\right)^{1/p}=\varepsilon^{1/p}$$ and $$\lVert f\rVert_p\lVert g\rVert_p=\varepsilon^{2/p}.$$


0

The main problem is that $v$ is not even well-defined. Indeed, the function $u$ and $$\tilde u(y) = u(y) + \begin{cases} 1 & \text{if } y = x + t \, h \text{ for some } t \in [0,1]\\ 0 & \text{else}\end{cases}$$ belong to the same equivalence class in $L^2(\mathbb R^n)$. However, both functions would be mapped to a different $v$.


1

Consider this counterexample. Let $u:\mathbb{R}^2\to\mathbb{R}$ s.t. $$ u(x,y)=\begin{cases} \frac{1}{\sqrt{x^2+y}} & (x,y)\in [0,1]\times[0,1]\\ 0 & \text{otherwise} \end{cases} $$ We can show this function is in $L^2(\mathbb{R}^2)$ (according to wolfram alpha, $$ \int \vert u(x,y)\vert^2 dxdy=\int_0^1 \int_0^1 \frac{1}{x^2+y}dxdy \approx 2.26394 ...


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 ...


1

Assume that $p=\alpha=2$ and that we work on the unit interval. If the sequence $\left(u_n^2\right)_{n\geqslant 1}$ converges weakly to $u^2$ in $\mathbb L^1$, then using the fact that the constant function equal to $1$ belongs to $\left(\mathbb L^1\right)'$, we derive that $$\tag{*}\lVert u_n\rVert_2\to \lVert u\rVert_2.$$ If $\left(u_n\right)_{n\geqslant ...


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:


1

No. Here is a counterexample: $g(x)=f(x+a)$ for $a\in\mathbb{R}$, $a\ne0$.


1

My try: Let $X=\cup_N X_N$ with $\mu(X_N)<\infty$, $X_N\subset X_{N+1}$, and define $E_N=(X_N\times X_N)\cap\{(x,y)\mid|k(x,y)|\le N\}$. Notice that $X\times X=\cup_N E_N$. We define a sequence of integral operators $K_N$ with kernel $k_N = \chi_{E_N}k$, where $\chi_{E_N}$ is the characteristic function of $E_N$. The operators $K_N$ are bounded, in fact, ...


1

As Wikipedia points out, there are several results called Riesz (representation) theorem. A common feature of some of them is that on some spaces, every linear functional is integration against something. The result that the author uses here is the representation of linear functionals on $L^q$: they are integrals against $L^p$ functions. More precisely, ...


1

Unless I've missed something, the authors made a little mistake there, but the result is of course true. Lemma 2.18 states that there exists $h\in K$ such that $$ \operatorname{Re}\int (k-h)u \leq 0 $$ for all $k\in K$. This can be rewritten into $$ \operatorname{Re}\int ku \leq \operatorname{Re}\int hu $$ for all $k\in K$. Not having $0$ on the right hand ...


0

To construct functions that negate these statements, have in mind $\{x_n\} \subset R$ such that one of the $\Sigma x_n^p$ and $\Sigma x_n^q$ converges but not the other. On sequences of subsets with the properties assumed above, define $f$ piecewise. You will still have a measurable and quite well-behaved function, and you must be able to find bounds ...


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 ...


0

Consider a bijective function $\varphi\colon (0,1)\to\mathbb R$ with continuous derivative, and such that $\lim_{x\to 0}\varphi(x)=-\infty$ and $\lim_{x\to 1}\varphi(x)=+\infty$. Then define a map $$\Psi\colon \mathbb L^p(\mathbb R)\to \mathbb L^p((0,1)), \Psi\colon\left(x\mapsto f(x)\right)\mapsto \left(x\mapsto f\left(\psi(x)\right)\varphi'(x)\right).$$ ...


3

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 ...


-1

It seems like a simple application of Dominated Convergence Theorem. From assumption, we have that $|f_i|^2\longrightarrow |f|^2$ a.e on $[0,1]$. But \begin{equation*} \int|f_i|^2\, dm\le \left(\int|f_i|^4\, dm\right)^{\frac{1}{2}}\le 1 \end{equation*} where the first inequality follows from Holder inequality. Thus, DCT implies that \begin{equation*} ...


0

Observe that $$ \int_E |f|^p =\int_E 1|f|^{p} $$ Now what you want to do is to use the Holder inequality, so that, using as exponent for f $ \quad \frac{p}{r} \ \ $ $$ \int_E |f|^r \leq \left( \int_E 1^{} \right)^\frac{p-r}{r} ||f ||_p ^{p}$$


1

As you wrote in a comment we have the Hölder inequality: If $h\in L^p(\mathbb{R})$ and $k\in L^{p'}(\mathbb{R})$, where $1/p+1/p'=1$ and $1\leq p,\,p'\leq +\infty$ then $$ \int_\mathbb{R} |h(x)k(x)|dx\leq \left(\int_\mathbb{R} |h(x)|^pdx\right)^{1/p}\left(\int_\mathbb{R} |k(x)|^{p'}dx\right)^{1/p'} $$ In our case we look at $f\in L^1(\mathbb{R})$ and ...


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 ...


1

Choose a continuous function $f$ so $f(0)=0$ and $||1-f||_2<\epsilon$. Choose a polynomial $q$ so $||f-q||_\infty<\epsilon$. Let $p=q-q(0)$; then $||p-q||_\infty<\epsilon$. So $||1-p||_2\le 3\epsilon$.


1

What you need to show is that $\|Ix\|_q \le \|x\|_p$ for all $x \in \ell_p$ and that the ratio $\dfrac{\|Ix\|_q}{\|x\|_p}$ can be made arbitrarily close to $1$. The second point is easy. If a sequence $x$ has exactly one nonzero term then $\|Ix\|_q = \|x\|_p$. On the other hand if you write $\|x\|_\infty$ for the maximum of the terms of a sequence $x \in ...


1

Note The theorem below is actually trivial without the lemma. The lemma seems interesting as well; this will be rewritten as a three part tfae soon. Theorem Suppose $1\le p<\infty$ and $X$ is a subspace of $\ell^p$. Then $X$ has some set of $e_j$ for a basis if and only if $X$ is monotone, meaning that if $x\in X$ and $|y(n)|\le|x(n)|$ for all $n$ then ...


1

The answer is no. A subspace of $\ell_p$ need not have a basis at all. However you don't need that. Take the subspace spanned by $e_i+e_{i+1}$ for $i=2,4,6,\ldots$.


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 ...


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 ...


0

Let $U\subset\ell^\infty$ be the subspace of sequences $x=(x_k)_{k\geq1}\in\ell^\infty$ with compact support, and define $\phi:\>U\to{\mathbb R}$ by $$\phi(x):=\sum_{k=1}^\infty x_k\ .$$


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.


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 ...


0

Better might be to show that every $f \not \in S$ has an open neighborhood $U_f$ where $U_f \cap S = \varnothing$. Then since the complement of $S$ (the union of all the $U_f$) is open, $S$ is closed. You're approximately there, I'm just not sure you're seeing your sentences as going in this direction.


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

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 = ...



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