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1

Note that $$f_x = \frac{(1 + x^2) \frac 1 2 x^{-1/2} - x^{1/2} \cdot 2x}{(1 + x^2)^2} = \frac{\frac 1 2 x^{-1/2} - \frac 3 2 x^{3/2}}{(1 + x^2)^2}$$ For large values of $x$, we have the estimate (where $\lesssim$ means "up to some constants") $$|f_x(x)| \lesssim \frac{x^{3/2}}{x^4} = x^{-5/2}$$ For small values of $x$, we have the similar estimate ...


6

Hint: you can show that $l^1$ is separable, and $l^\infty $ is not.


0

The following gives the main idea. \begin{eqnarray*} |F(k)| &\leqslant &\frac{1}{(1+k^{2})^{2}}\Rightarrow F(k)\in L^{1}\cap L^{\infty } \\ |kF(k)| &\leqslant &|k|\frac{1}{(1+k^{2})^{2}}\Rightarrow H(k)=kF(k)\in L^{1}\cap L^{\infty } \end{eqnarray*} It follows that $g(x)\in L^{1}$ and $g(x)\in L^{2}\cap L^{\infty }$ so $% g(x)\in L^{1}\cap ...


1

The inclusion $L^2\subset L^1_w$ does not hold: take $f(x):=\frac 1{\sqrt x\log x}\chi_{(1,\infty)}(x)$. The converse reduces to ask whether $g\in L^1$ implies $x\mapsto xg(x)^2\in L^1$. Define $$g(x):=\sum_{j=1}^\infty c_j\cdot\chi(j^2-a_j,j^2+a_j)(x),$$ with the $a_j$ small enough. Then $g$ is integrable if and only if $\sum_{j=1}^\infty c_ja_j$ is ...


0

By a change of variables $u=\frac{y}{x}$, $$\int_0^a e^{\frac{p}{x}} d\lambda(x)=y\int_{y/a}^\infty \frac{e^u}{u^2} d\lambda(u)$$ But $e^{u}/u^2>\frac{1}{u^2}+\frac{1}{u}+\frac{1}{2}$ is not integrable at $\infty$.


1

This can be generalized: this is a notion called uniform integrability. It gives nice convergence properties in mesure theory (in particular a generalization of the dominated convergence theorem).


1

I hope I didnt miss something, but this should work, For $p=1$ the inequality trivialy holds and notice that $\forall p>0$ we have, $$|a-b|^p\leq|a+b|^p$$ We have two cases to consider, (1) $p>1$: Note that $h(x)=x^p$ is convex and monotone for $p>1$. Hence, ...


0

Given a measurable space $X$ equipped with a measure $\mu$, a function $f : X \to \mathbb{C}$ which is defined almost everywhere (that is, up to a set of measure $0$) is said to be an element of $L^1$ if $$\int_X |f| d\mu < \infty$$ More properly, we have to identify functions which are equal almost everywhere, so the elements of the Lebesgue space ...


0

As @John pointed out in the comment, your apporach may not work, because $u'_n$ does not need to be dounded in $C^0$. One way to approach this problem is the following. We will use Theorem 4.26. from Brezis book. To this end, extend all functions $u_n$ to $\mathbb{R}$ by using Thereom 8.6. of the same book and note that the extended sequence still bounded. ...


1

An $L^1$ functional from a space $X$ to $\mathbb{R}$ is an $\mu$-measurable function such that $$ \int_{X} |f|\,d\mu < \infty. $$


1

Yes, your proof is good. In terms of writing, you start using a fixed $x\in\ell^1$ without saying so. Also, the sentence that says "For that we will show..." doesn't really make sense; I understand it because I know how to prove the density, but otherwise it looks hard to understand.


2

Write $s = tp+(1-t)q$. Consider $1/p'+1/q'=1$. $$ \int |f|^{s} = \int |f|^{tp+(1-t)q} \le \left\{\int |f|^{tpp'} \right\}^{1/p'} \left\{\int |f|^{(1-t)qq'} \right\}^{1/q'} $$ Now you can choose $$ tpp'=p, (1-t)qq'=q\iff p'=\frac 1t, q'=\frac 1{1-t} $$ because $$ \frac 1{p'}+\frac 1{q'} = t+(1-t)=1 $$ Then $$ \int |f|^{s} \le ||f||_p^{p/p'} ||f||_q^{q/q'} ...


1

Hint: It follows as a corollary of the following statement: Let $\{f_i:i\in I\}$ be a family of functions with $f_i\in L^{p_i}(\Omega)$ and $\dfrac{1}{p}=\sum \dfrac{1}{p_i}\leq1$. Then $\prod f_i\in L^p(\Omega)$ and $$\|\prod f_i\|_{L^p(\Omega)}\leq\prod\|f_i\|_{L^{p_i}(\Omega)}$$ This is the interpolation inequality (H. Brezis "Functional Analysis, ...


1

Here's an idea that might be helpful: that $L^p$ measures spikiness or broadness of a function in some sense. In particular, $L^{\infty}$ consists of functions with no spikes whatsoever, while $L^1$ functions can't be too broad; the higher $p$ is, the more control you have over spikes, and less control over broadness. Since $f \in L^p \cap L^q$, we've ...


1

If a function $u$ is actually a radially symmetric function, then we have the integration formula $$ \int_{B(0,R)} u(x)\, dx = |\mathbb{S}^{n-1}| \int_0^R u(r)r^{n-1}\, dr, $$ where $n$ is the dimension of the space where $u$ is defined. In your case, $u(x)=|x|^{-\lambda}$, and it is straightforward to apply the formula and check the integrability of $u$.


2

Given $f \in L^\infty$, you can find a sequence $f_n \in C_c^\infty$ such that $\|f_n\|_\infty \le \|f\|_\infty$, and $f_n \to f$ in $L^1$, that is, $\|f-f_n\|_1 \to 0$.


2

If $(g_n)$ is a sequence in $C_c^{\infty}$ and $\|g_n-f\|_{L^p}\rightarrow0$, then $(g_n)$ is a Cauchy sequence and the limit $g\in L^p$ exists. It can easily be shown that $f=g$ almost everywhere.


1

EDIT: the fact that the sequence is unbounded for $y = 0$ is not a problem since we don't care about sets of measure $0$. Clearly $$\lim_{n \to \infty} \frac{1}{1 + x^2}f\Big(\frac{x}{\sqrt{n}}\Big) = \frac{1}{1 + x^2}f(0).$$ Moreover we can find $N$ such that if $n \ge N$ then $\Big|f\Big(\frac{x}{\sqrt{n}}\Big) - f(0)\Big| \le 1$, then in particular ...


1

$A$ is closed, but not open. The sequence $x=(1,1,1,...)$ is not an interior point. If $(x_n)$ is a convergent sequence in $A$ with the limit, then it also converges pointwise. But since $0\le x_{n,m}\le 1$, we also have $0\le \lim \limits_{n\to \infty} x_{n,m}\le1$, so $\lim \limits_{n\to \infty}x_n\in A.$ $B$ is open, but not closed. If $f\in B$, not that ...


2

$A$ is not closed. Take any function $f\in L^1$ such that $f$ is not in $L^2$, then approximate it by functions in $A$. I.e. $f(x) = \chi_{(0,1)}(x) x^{-1/2}$, then $f\in L^1$, $f\not\in L^2$. Define $f_n(x) = \min(n,f(x))$. Then $f_n\to f$ in $L^1$, $\|f_n\|_{L^2}\to\infty$. As to your second question: sets like $\{f\in L^1: \ f\ge 0 \}$ are closed as ...


3

From the inclusions of sets $$\ell_p\subset c_0\subset \ell_\infty,$$ where $c_0$ denotes the set of convergent sequences and the separability of $(c_0,\lVert \cdot\rVert_\infty)$ is separable we conclude that $(\ell_p,\lVert\cdot\rVert_\infty)$ is separable. $\ell_p$ endowed with the supremum norm is not a closed subspace of $\ell_\infty$ because its ...


-2

Generally, to prove that $A=B$ for sets $A,B$ one shows that if $a \in A$ then also $a \in B$ (i.e. $A \subseteq B$) and also if $b \in B$ then also $b \in A$ (so $B \subseteq A$). Let $A = L^2(0,\pi)$ and let $B = \mathrm{span}\left(\left\{\sin kx\right\}_{k=0}^\infty\right)$. How do you proceed to apply this to your problem?


1

Unfortunately, a sequence can be weakly convergent without being pointwise convergent $(f_n(x))=(\sin(n\pi x))$ on the unit interval (which is not convergent except at $0$). However we can show that for each simple function $g$ (linear combination of characteristic functions of measurable sets), $$\lim_{n\to +\infty}\int_{(a,b)} f_ngdx= \int_{(a,b)} ...


1

Consider the unit interval $[0,1]$ and $$f_\epsilon(x) = \frac 1 {\sqrt x} \chi_E = g_{\epsilon}(x)$$ where $E = [\epsilon, 1]$ and $\chi_E$ is a characteristic function. Then $f_{\epsilon}g_{\epsilon} \in L^1$, and we have $$\|f_{\epsilon}\|_1 = \int_{\epsilon}^1 \frac{1}{\sqrt x} dm(x) \le \int_0^1 \frac{1}{\sqrt x} dm(x) = 1$$ and likewise for ...


4

In fact, we can derive that $\,\dim S \le c^2$. Let me describe the proof as it is very elegant. Assume that $v_1,\ldots, v_n\in S$ are orthonormal functions, i.e., $\int_0^1 v_iv_j\,dx=\delta_{ij}$, and for a fixed $a=(a_1,\ldots,a_n)\in\mathbb R^n$ define $\varPhi_a :\mathbb R^n\to \mathbb R$ as $$ \varPhi_a(x)=\sum_{j=1}^n a_jv_j(x). $$ Then $$ ...


2

Here is a counterexample. Let $\phi\colon\mathbb{R}\to\mathbb{R}$ be $C^\infty$ supported on $[0,1]$, positive with $\int_0^1\phi(x)=\int_0^1\phi(x)^2\,dx=1$. Let $$ f_n(x)=\sum_{k=1}^n2^{k}\phi(2^{2k}(x-k)). $$ Then $$ \int_{\mathbb{R}}f_n(x)\,dx=\sum_{k=1}^n2^{k}2^{-2k}<1. $$ Moreover $f_n(x)\le f_{n+1}(x)$ for all $x\in\mathbb{R}$. It follows that ...


3

No, you can't expect. You can approximate any $f\in L^1$ by functions $f_n$ in the Schwartz space $\mathscr S$ (even by smooth functions with compact support). Then $\hat f_n \in \mathscr S \subseteq L^1$ but there is no reason to expect convergence of $\hat f_n$.


8

You can apply a theorem of Grothendieck to the closure of $S$ in $L^2$ which is (as you show) contained in $C([0,1]) \subseteq L^\infty$. Grothendieck's theorem says that every closed subspace of $L^p(\mu)$ (where $\mu$ is a probability measure on some measurable space and $0<p<\infty$) which is contained in $L^\infty(\mu)$ is finite dimensional. A ...


4

The statement is false, as I discovered here. Since not everybody has access to the paper, let me provide a summary of the argument: Let $R=\mathbb C[t]$, $L$ the left shift operator, and view $\ell^p$ as an $R$-module by defining $t\cdot x=Lx$. Let $X=\sum \ker L^i \subset \ell^p$ be the subspace of eventually-zero sequences. Lemma: Given a PID $P$, a ...


3

There is no reason to expect that the $g_n$ converge, the analogous implication does also not hold in $\mathbb{R}$. For an - admittedly boring - explicit example, take $f_n = \chi_{[0,2]}$ for all $n$ (where $\chi_A$ is the characteristic function of the measurable set $A\subset \mathbb{R}$), and $g_n = \chi_{[n,n+1]}$. Evidently the constant sequence $f_n$ ...


0

We are given that $\lVert f_n\rVert_p$ converges to $0$. Good. But it can converge be arbitrarily slowly. However, given a sequence $(c_k)_k$ of real numbers, we can find a subsequence $(f_{n_k})_k$ such that $\lVert f_{n_k}\rVert_p\leqslant c_k$ for each $k$. In particular, we obtain for each $k$, $$\mu\{x,|f_{n_k}(x)|\geqslant k^{-1}\}\leqslant kc_k.$$ ...


0

In a general context, note that $L^p$ convergence $\implies$ convergence in probability $\implies$ almost sure convergence (up to an extraction!). edit: for a simple example, consider $$ f(x) = 1_{nx<1} \text{ if $n$ is even}\\ f(x) = 1_{n(1-x)<1} \text{ if $n$ is odd} $$and the subsequence $\phi(n) = 2n$.


1

If you are working in $\mathbb{R}$: Without loss of generality assume that $f,g \geq 0$. For each $n \in \mathbb{N}$ let $A_n \subseteq \mathbb{R}$ such that $0<\mu(A_n)<\infty$ and $f>n$ on $A_n$, where I use $\mu$ to denote Lebesgue measure on $\mathbb{R}$. Let $g_n:= 1_{A_n}/\mu(A_n) \in L^1$. Then $$\|fg_n\|_1 = \frac{1}{\mu(A_n)}\int_{A_n} ...


0

If $L^{1}=L^1(\Omega)$ where $\Omega$ is $\sigma$-finite, then $(L^{1})^*=L^\infty$, and if your claim didn't hold, there would be an $M>0$ such that for all $g\in L^1$, $\|fg\|_{L^1}<M$, and so $g\mapsto\int fg$ would be in $(L^1)^*$, which is impossible. So under the assumption that your measure space is $\sigma$-finite, we can actually find a ...


2

Hint: given any continuous function on $[0,1]$, you can approximate it in the $L^1$ norm by continuous functions that vanish at $0$.


1

Just as you suspect, $\left\|\cdot\right\|_{p,q}$ is the norm $L_p \rightarrow L_q$. In this context $\left\|\cdot\right\|_{2}$ is just $\left\|\cdot\right\|_{2,2}$ and so on.


3

Note that for every $t\ge 0$, we have that $$ t^h\le t^p+t^q. $$ To see this consider the possibilities $t\le 1$ and $t>1$. Hence $$ \lvert f(x)\rvert \le \lvert f(x)\rvert^{p/h}+\lvert f(x)\rvert^{q/h}=g_1(x)+g_2(x)=g(x), $$ and thus $$ \|f\|_h\le \|g\|_h\le \|g_1\|_h+\|g_2\|_h=\left(\int_E g_1^h\right)^{1/h}+\left(\int_E ...


1

Hint 1 I believe $$f \in L^p(0,\infty) \Leftrightarrow \int_0^\infty |f(x)|^p dx < \infty.$$ So you need to find such a $p$ that $$ \int_0^\infty \frac{dx}{\left( x^\alpha + x^\beta \right)^p} < \infty. $$ How would you proceed to do that? Can you get an idea about how to evaluate that integral? Hint 2 A possibly useful transformation is $$ ...


1

If $K$ consists of finitely many atoms and a set of measure zero, then your set of functions is finite-dimensional, and compactness follows from the compactness of closed bounded subsets of $\mathbb R^n$. But then $X$ is not such an interesting space to look at. If $K$ contains a set $A_0$ of positive measure with no atoms, then no. By a theorem of ...


1

This is a very standard and yet elegant proof. We shall create a bounded linear functional on $\ell^\infty(\mathbb N)$, which is not realized by any element of $\ell^1(\mathbb )N $, using the Hahn-Banach. Let first $c(\mathbb N)$ be the set of converging sequences. Clearly $c(\mathbb N)$ is a closed subspace of $\ell^\infty(\mathbb N)$. Let ...


3

If $(x_k)$ is a sequence of vectors in $\ell_q$, each with $p$-norm at most one, that converges to $x$ in $\ell_q$, then $(x_k)$ converges to $x$ coordinatewise. For each $k$ and $n$, $\sum\limits_{i=1}^n |x_k(i)|^p\le 1$. Fix $n$ and let $k\rightarrow\infty$ to deduce that $\sum\limits_{i=1}^n |x(i)|^p\le1$ for each $n$. Now let ...



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