# Tag Info

1

You want $i : L^p \cap L^r \to L^q$ to be continuous when you equip the domain with the $L^p$ norm and the domain with the $L^q$ norm. Since it is linear, it is equivalent for it to be bounded, i.e. $$\| f \|_q \leq C \| f \|_p$$ for all $f \in L^p \cap L^r$. This is actually not true. Here's a proof. Suppose $f \in L^p \setminus L^q$, without loss of ...

1

Hint: What we need to prove to prove continuity of such inclusion map is that there exists a constant $C>0$ such that, for $f \in L^p \cap L^r$, $$\|f\|_q \leq C \|f\|,$$ where $\|f\|=\|f\|_p + \|f\|_r$ is a norm on $L^p \cap L^r$.

0

No. If the norm were induced by an inner product it would satisfy the parallelogram law $$||f+g||^2+||f-g||^2=2(||f||^2+||g||^2).$$Simple examples show this is not so (for example, the characteristic functions of two disjoint sets).

0

Hint: if a norm is derived from an inner product in that way, then the inner product is uniquely determined by the norm and there is an explicit algebraic expression for (f,g) in terms of ||f||, ||g||, ||f+g|| and ||f-g||.

1

Hint: take a look at the parallelogram law.

1

3: Yes.

1

No. Yes, if the space has finite measure. Not sure, maybe somebody else can answer this. Well this is hard to answer, but one missing "concept" is $L^p$ convergence implies convergence in probability. You should also see how this relates to convergence in distribution - these are all topics covered in graduate courses in probability theory (and partly in ...

0

Multiplying $f$ by a constant $a>0$ results in all norms multiplied by $a$. Multiplying $\mu$ by a constant $b>0$ results in $L^p(\mu)$ norm multiplied by $|b|^{1/p}$. So, you want $$ab^{1/p_0} = 1/\|f\|_{p_0},\qquad ab^{1/p_1} = 1/\|f\|_{p_1}$$ and this system can be solved for $a$ and $b$.

1

Hint: If $|f|\le 1,$ then $|f|^{p_2} \le |f|^{p_1}.$

2

Let $f \in L^{p_1}(E)$ bounded. Then there exists $M\geq 0$ such that $\sup_{x \in E} |f(x)| \leq M$ and $\|f\|_{p_1}^{p_1}=\int_E |f(x)|^{p_1}\,dx < \infty$. Case 1: $p_2<\infty$ For $p_2 > p_1$, we have \begin{align} \|f\|_{p_2}^{p_2}& =\int_E |f(x)|^{p_2}\,dx \\ &= \int_E |f(x)|^{p_2-p_1}|f(x)|^{p_1}\,dx \\ & \leq \left(\sup_{x ...

1

A short answer, if $\sqrt{\rho} \in H^1(\mathbb{R})$, then by the Sobolev embedding ( https://en.wikipedia.org/wiki/Sobolev_inequality#Sobolev_embedding_theorem with $1/p^* = 1/6 = 1/2 -1/3 = 1/p - 1/n$), we have $\sqrt{\rho} \in L^6(\mathbb{R}^3)$, which then results in $\rho \in L^3(\mathbb{R}^3)$. You can even get estimates on the norm this way. edit: 36 ...

2

By Sobolev embeeding, from $\sqrt\rho\in H^1(\mathbb R^3)$ it follows that $$\sqrt \rho \in L^6(\mathbb R^3),$$ which implies $$\rho \in L^3(\mathbb R^3).$$

1

Hint: Use Holder inequality to show that $A_f$ is bounded

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You can do this change of variables : u = $\log{1/x} \implies x = e^{-u}$ So, to calculate the p-norm, consider the following integral $\int_0^{1}\log(\frac{1}{x})^pdx = \int_{0}^{\infty} u^p e^{-u} du$ The integral in the r.h.s is the well-known gamma function $\Gamma(p+1)$, which converges for the values of p that we are interested. For the ...

1

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Ilya Bogdanov below. $\def\sign{\mathop{\rm sign}}$First of all, it is enough to prove the statement when $p=u/v$ is rational, $u$ is even and $v$ is odd (such numbers are dense on the real line). We need this to simplify the last argument. Let the ...

1

For any $a>0,\lim_{x\to 0^+}x^a\ln (1/x) =0.$ So given $0<p<\infty,$ we have for small positive $x$ $$x^{1/(2p)}\ln(1/x) < 1 \implies x^{1/2}(\ln(1/x))^p < 1 \implies (\ln(1/x))^p < x^{-1/2}.$$ Since $\int_0^1 x^{-1/2}\, dx < \infty,$ we have the result.

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Make a simple change of variables, $u=x^{-1}$ so that $du = -x^{-2}dx\iff -u^{-2}du = dx$ Then you get $$\int_\infty^1 -{\log u\over u^2}\,du=\int_1^\infty {\log u\over u^2}\,du$$ And this is easily verified to be integrable by directly comparing with $$\int_1^\infty {du\over u^{3/2}}$$ Since we know $$\lim_{x\to\infty} {\log x\over \sqrt{x}}=0$$ So ...

2

If $\mu(X)=\infty$, the result is not true in general. Let $X=[1,\infty)$ with $\mu$ Lebesgue measure and let $f_n$ be the characteristic function of the interval $[n,2\,n]$. Then $f_n\in L^p$, $\|f_n\|_\infty=1$ for all $n$ and $f_n(x)$ converges point wise to $f(x)=0$ for all $x\in X$. Let $g(x)=1/x$. Then $g\in L^q$ for $q>1$. We have $$\int_X ... 2 The Fourier transform is a linear map, you only have to check that$$\forall f\in L^1(\mathbb R), \mathcal F(f)=0 \Rightarrow f=0.$$Let f\in L^1(\mathbb R) such that \mathcal F (f)=0. Hence \mathcal F(f) is L^1(\mathbb R) since its the zero function and therefore its Fourier inverse exists. So$$f=\mathcal F^{-1}(\mathcal F(f))=\mathcal ...

2

For any $p \geq 1$, we have $$|x+y|^p \leq 2^p (|x|^p+|y|^p),$$ and therefore \begin{align*} \mathbb{E}(|X_n-X|^p \mid \mathcal{F}) &\leq 2^p \mathbb{E}(|X_n|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}) \\ &\leq 2^p \mathbb{E}(|X|^p \mid \mathcal{F}) + 2^p \mathbb{E}(|Y|^p \mid \mathcal{F}). \end{align*} This shows that ...

0

Work from the other end, using Continuity of $L^1$ functions with respect to translation. Weak* continuity in $L^\infty$ means that for every $g\in L^1$, $$\int (\tau_{h}f)g\to \int fg,\qquad h\to 0$$ But $\int (\tau_{h}f)g= \int f(\tau_{-h}g)$, by a change of variables. Also, $$\left|\int f(\tau_{-h}g) - \int fg\right| \le \|f\|_\infty \|\tau_{-h}g ... 2 Later hint, p=2: Let \epsilon >0. For 0<x<\epsilon we have$$|f(\epsilon)-f(x)| \le \int_x^\epsilon |f'(t)|\, dt \le (\int_x^\epsilon|f'(t)|^2t\, dt)^{1/2}(\int_x^\epsilon t^{-1}\, dt)^{1/2}\le (\int_0^\epsilon|f'(t)|^2t\, dt)^{1/2}(\int_x^\epsilon t^{-1}\, dt)^{1/2}.$$Note the first integral on the last line \to 0 as \epsilon \to ... 1 I assume by C:= conv^{\ast}(e_i) you mean the closed convex hull of the \{e_i\}. Let \varphi \in (\ell^1)^{\ast} denote the linear functional$$ \varphi((x_n)) := \sum_{n=1}^{\infty} x_n $$Then for any z \in \text{conv}(\{e_i\}), we have$$ \varphi(z) = 1 $$Since C is norm closed, it follows that \varphi \equiv 1 on C. In particular, 0\notin ... 1 Consider$$ f_n := n\chi_{[0,1/n]} \in L^p[0,1] $$Then if n<m,$$ \|f_n - f_m\|_p^p = \int_0^{1/m} (m-n)^pdt + \int_{1/m}^{1/n} n^pdt \geq \frac{(m-n)^p}{m} $$Hence if p\geq 1, this sequence cannot have a convergent subsequence. 1 One way to do this is to produce a sequence (u_n)_n of functions in the unit ball of L^p(0,1) such that \|u_i - u_j \|_{L^p} \ge c for some c > 0, all i \ne j. Then no subsequence can converge. You can get such a sequence by defining$$ u_j(x) = \begin{cases} 2^{k/p}\quad (\frac{\ell}{2^k} \le x \le \frac{\ell+1}{2^k}\\ 0 \quad ...

2

$U$ is some other normed vector space. In this case $L^2([0,T];U)$, sometimes lazily written as $L^2(0,T;U)$, consists of functions $f$ from $[0,T]$ to $U$ such that $\int_0^T \| f(t) \|^2 dt<\infty$, where $\| \cdot \|$ is the norm on $U$. This notation is used in, for instance, Partial Differential Equations by Evans. Most commonly the $U$ in question ...

2

I'm not sure about better, but here is an alternate solution based on Holder's inequality: since $p_1 < p < p_2$ you have $\frac 1{p_2} < \frac 1p < \frac 1{p_1}$ so there exists a constant $0 < \alpha < 1$ with $\frac 1p = \frac{\alpha}{p_1} + \frac{1-\alpha}{p_2}$. Then $1 = \frac{p\alpha}{p_1} + \frac{p(1-\alpha)}{p_2}$ and by Holder's ...

1

This holds iff $w \ge 0$ a.e. In this case, the functional $u \mapsto \int_0^T w(t) \, \|u(t)\|_{L^2(\Omega)} \, \mathrm{d}t$ is convex and continuous, hence weakly lower semicontinuous. If $w < 0$ a.e. on a subset $I$ of positive measure, take your favorite sequence $\{v_n\} \subset L^2(\Omega)$ with $v_n \rightharpoonup 0$, but $v_n \not\to 0$ in ...

2

By divergence theorem $$\int_{\partial \Omega }u\cdot n \ \phi=\int_\Omega div(u\phi) = \int_\Omega \phi div(u) + \nabla \phi \cdot u.$$ This gives $$\left|\int_{\partial \Omega }u\cdot n \ \phi\right|\le \|\phi\|_{L^2(\Omega)}\|div(u)\|_{L^2(\Omega)} + \|\nabla \phi\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \le \|\phi\|_{H^1(\Omega)}\|u\|_{H(div,\Omega)}.$$

0

The multiplication operation on $L^1(\mathbb{R})$ is convolution. If you choose $f \in L^1$ such that $f \ne 0$ and $\hat{f}(\xi)=0$ for a fixed, given $\xi\in\mathcal{R}$, then $h=f\star g = g\star f$ also has the property that $\hat{g}(\xi)=0$ (for this you need the Fourier transform and $\widehat{f\star g}=C\hat{f}\hat{g}$ where $C$ is a constant. And, if ...

3

Let $\lambda$ denote the Lebesgue measure. If $\lambda(A) = \infty$, then both $a_f$ and $a_g$ are zero. Indeed, for $a>0$, $$\lambda(\{x\in A: |f(x)|> a/2\})\le 2 a^{-1}\int_{A} |f(x)|\, dx<\infty,$$ therefore, $$\int_{A} |f(x)-a|\, dx\ge \frac{a}2\lambda(\{x\in A: |f(x)|\le a/2\}) = \infty.$$ (Indeed, $|z-a|\ge \frac{a}2\mathbf{1}_{|z|\le ... 1 A small remark. Note that both series converge naturally in$L^2$, considering that the basis functions are elements of the Hilbert space (and thus, not defined pointwise). We could additionally have almost everywhere pointwise convergence, but this requires proof: it is not direct from Hilbert space rules. Answer. As$f_0 = 0$, there is no constant ... 0 Look at what each term in the sum is:$\left(\frac{1}{|\mathcal{Q}_i|}\int_{\mathcal{Q}_i}fdx\right)\chi_{\mathcal{Q}_i}$is a step function taking the average value of$f$over$\mathcal{Q}_i$in$\mathcal{Q}_i$and zero everywhere else--in particular, zero on all other$\mathcal{Q}_j$s. Perhaps for ease of notation you should write ... 2 This is not true in general. Let$X = \ell^2$, and$T \in L(\ell^2)$given by $$Tx = \left(\frac{x_n}{n+1}\right)_n$$ Then$T$is one-to-one, but for$g = e_n \in \def\ran{\operatorname{im}}\ran T$we have that the inverse image is$f = (n+1)e_n$, and $$\|f\| = \|(n+1)e_n\| = (n+1)\|e_n\| \stackrel ?\le C \|e_n\|$$ Hence, such a$C$cannot exist. 3 A direct proof, from which a proof by contradiction can be derive if necessary. Let$A_n=\{x \in \mathbb R \ : \ \vert f(x) \vert^p \ge n\}$for$n$integer and$b = \int \vert f(x) \vert^p dx$. We have $$0 \le n \mu(A_n) \le b \tag{1}$$ where$\mu$is the measure. Hence$\mu(A_n) \le \frac{b}{n}$for all$n \in \mathbb N$. $$C=\{x \in \mathbb R \ : \ ... 1 I think you're not on the right track. You already pointed out some of the problems with your approach. Let me try to help you see how to tackle such a problem. First of all it helps to assume that f is a Schwartz function. In fact a bit more than your claim is true. Claim. If for some 1\le p,q\le \infty, there exists a constant ... 0 The closed unit ball in a normed space is closed and bounded but compact if and only if the space is finite dimensional. You can find a proof that uses Riesz's lemma here. 1 As suggested in a comment by John Dawkins, you need the Cauchy-Schwarz inequality:$$ f(x)^2 \le \left(\int_0^x |f'(t)|\,dt\right)^2\le \int_0^x 1\,dt \int_0^x |f'(t)|^2\,dt = x \int_0^x |f'(t)|^2\,dt $$Here the factor \int_0^x |f'(t)|^2\,dt tends to zero as x\to 0, and the conclusion x^{-1}f(x)^2\to0 follows. 0 For completeness, a counterexample for p>q: pick s\in (1/p,1/q) and consider the sequence x_n=1/n^s. This sequence is in \ell^p but not in \ell^q. 3 For any sequence y\in c_0 we have$$ \|x-y\|_\infty = \sup_n|x_n-y_n|\ge \limsup_{n\to\infty} |x_n-y_n|=\limsup_{n\to\infty} |x_n| $$This gives a lower bound on the distance. To get the matching upper bound, let y_n=x_n when n\le N, and y_n=0 otherwise. This is an element of c_0, and$$ \|x-y\|_\infty = \sup_{n>N}|x_n| ... 4 $$\int_E \lvert f \rvert^p \cdot 1 \leqslant \left(\sup_E \lvert f \rvert^p \right) \int_E 1 = \lVert f \rVert_{\infty}^p m(E),$$ since$\lVert \lvert f \rvert^p \rVert_{\infty} = \lVert f \rVert_{\infty}^p$. Therefore, taking$p$th roots, $$\lVert f \rVert_{p} \leqslant m(E)^{1/p}\lVert f \rVert_{\infty}$$ It has to be this because the equality case ... 0 If$f_n$converges to$f$uniformly (i.e.$\sup_{x \in I}|f_n(x)-f(x)| \to 0$), then $$\|f_n-f\|_2=\int_I |f_n(x)-f(x)|^2\,dx \leq m(I) \left(\sup_{x \in I}|f_n(x)-f(x)|\right)^2 \to 0,$$ so$f_n$converges to$f$in$L^2$. If$f_n$converges to$f$in$L^2(I)$, then, by Hölder inequality we have $$\|f_n-f\|_1=\int_I |f_n(x)-f(x)|\,dx \leq ... 0 On a finite measure space L^q convergence implies L^p convergence for any 1\leq p \leq q \leq \infty. Take \infty>q>p, so q/p > 1.This follows from Holder's inequality,$$ ||f||_p^p = \int |f|^p\,d\mu = \int |f|^p \cdot 1\,d\mu \leq \left(\int (|f|^{p})^{q/p}\,d\mu\right)^{p/q}\left(\int 1\,d\mu \right)^{1-p/q}= ||f||_q^p \mu(X)^{1-p/q} $$... 1 For x\in (0,1), define$$f(x) = \frac{1}{x(|\ln x|+1)^2}.$$Then f \in L^1(0,1), but f \notin L^p(0,1) for all p>1. 1 This is the translation semigroup (actually, it's a group because you're working on \mathbb{R} and not [0,\infty).) There are various ways of dealing with this problem directly. For example, to show that it is a C^0 group, it is enough to show contuity of T(t)f for f in a dense subspace \mathcal{M} of L^{2}(\mathbb{R}), such as the subspace ... 2 Since \frac{p}{q}+\frac{q-p}{q}=1, apply Holder's inequality.$$ \left(\int_X |f|^p\,d\mu\right)^{1/p}\leq \left[\left(\int_X |f|^{q}\right)^{p/q}\mu(X)^{(q-p)/q}\right]^{1/p}=\left(\int_X|f|^q\right)^{1/q}\mu(X)^{(q-p)/(pq)}.$$Note Holder's inequality only applies if the condition$\mu(X)<\infty$holds. The inequality you ask for will hold ... 2 A function in$\mathcal{S}$decays faster than the reciprocal of any polynomial at infinity and is bounded. (This is usually built into the definition.) So its magnitude is bounded by some$M_1$on the ball of some radius$R$centered at the origin, and by$M_2/|x|^N$outside this ball. Choose$N$sufficiently large (depending on the dimension and$p$) to ... 0 The answer is NO to both questions. Indeed, to give a counter-example, it is sufficient to show that there exists a bounded measurable function$f$such that, for all negligible sets$N$,$f|_{\mathbb R^d\setminus N}$is not continuous (in the subspace topology). Let$\mathbb Q^d=\{q_n\}_{n\in\mathbb N}$the$d$-uples of rational numbers. Consider ... 1 MickG, the answer to both of your questions is no (at least in the presence of measurability). Consider all bounded measurable functions which are supported on the hypercube$[0,1]^N$. If you assertions were true,$L_\infty[0,1]^N$would be Banach-space isomorphic to$C[0,1]^N$but the former space is non-separable. 0 Consider the function f in$R$constructed as follows: Let M is a big natural number. Then let$m${x$\in R$:f(x)=M}=$\frac{1}{M}$. (where m is the lebesgue measure in$R$). Then$m${x$\in R$:f(x)=M+1} =$\frac{1}{(M+1)^2}$. Carry on this type of construction i.e in general$m${x$\in R$:f(x)=M+k} =$\frac{1}{(M+k)^{k+1}}$. Let {x$\in R$:f(x)=M+k}=$E_k\$ ...

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