# Tag Info

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The statement is false, at least if $c$ is to be independent of $f$. Fix $p\ge1$ and $\epsilon>0$ and suppose $$c^{1/p}(\|f\|_\infty-\epsilon)\le\|f\|_p$$ holds for some constant $c$. Let $f_n$ be a triangle function with base on $[-1/n^{p+1},1/n^{p+1}]$ and height $n$, and zero otherwise. Then $\|f_n\|_\infty=n$ and $$\|f_n\|_p^p=2\int_0^{1/n^{p+1}}(n-n^{... 1 You cannot find a constant c independent of f. If you allow it to depend on both f and \epsilon but not on p then you can use the following argument: There exists an interval (c,d) \subset [a,b] such that |f(x)| >M-\epsilon for c < x<d. We then get \int_a^{b} |f(x)|^{p}dx \geq \int_c^{d} |f(x)|^{p}dx \geq (M-\epsilon)^{p} (d-c). So ... 2 Ok here's what I got. Let's clear up the relationship between n \text{ and } d. . If n < d then by Holder's inequality$$ ||u||_{W^{1,n}(\mathbb{R}^d)} \leq C(d,p)||u||_{W^{1,d}(\mathbb{R}^d)} $$which reduces to the case n = d. If  n > d then we may apply Morrey's Inequality to show that u \notin W^{1,n} here. Let's look -$$ ||u||_{C^{0,\...

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For each $A \in \mathscr{A}$, since $\int_A f_i dP$ converge, let $\mu(A)= \lim_{i \to \infty} \int_A f_i dP$. We can prove that $\mu$ is a measure (and since $\int_A f_i dP$ is bounded by $1$ for every A $\in$ $\mathscr{A}$, $\mu$ is in fact a probability). One way to prove that $\mu$ is countably additive is to apply Vitali–Hahn–Saks theorem. Now, it is ...

2

The unit ball of $L^{2}[0,1]$ is not compact. Let $(g_n)$ be sequence in it with no convergent subsequence. Let $f_n(x)=g_n(x)$ for $x \in [0,1]$ and $0$ for all other $x$. Then $(f_n)$ is a bounded sequence in $L^{2}(\mathbb R)$ and $(Tf_n)$ has no convergencet subsequence. Hence $T$ is not compact.

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Usually convolution product is defined in this way: $$(f \star g)(s):=\int_{-\infty}^{\infty} f\left(s-t\right)g\left(t\right)dt \tag{1}$$ In the case of "causal" functions (i.e., functions which are zero or are made zero for negative values of $x$), we can write, instead of $f(t)$ and $g(t)$: $$U(t)f(t) \ \text{and} \ \ U(t)g(t)$$ where $U$ is the ...

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Let $u \in W^{1,1}(0,T;X)$ for some Banach space $X$, i.e., $u' \in L^1(0,T;X)$. By the fundamental theorem of calculus we have in $X$ $$u(t)=u(s)+\int_s^t u'(\tau) d\tau,$$ for almost all $s,t \in (0,T)$. Taking the norm of $X$ $$\|u(t)\|_X = \left\|u(s)+\int_s^t u'(\tau) d\tau\right\|_X \leq \|u(s)\|_X + \int_s^t \|u'(\tau)\|_Xd\tau.$$ Taking the essential ...

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The volume of an $\ell_2^n$ ball of radius $r$ is $r^n\cdot\frac{\pi^{n/2}}{\Gamma(n/2+1)}$. The volume of the $\ell_{\infty}^n$ unit-ball, i.e., the cube $[-1,1]^n$, is $2^n$. A straight-forward application of Stirling's formula shows that $$\lim_{n\to\infty}\left(\frac{\pi^{n/2}}{\Gamma(n/2+1)}\right)^{1/n}\sqrt{n}=\sqrt{2e\pi}$$ Therefore, what you wrote ...

1

So, here I consider $1<p<\infty$ and write $\frac{1}{p}+\frac{1}{q}=1$ for some $1<q<\infty$. Next let $F\in \ell_p'$ i.e. $F$ is a bounded or continuous real linear functional on $\ell_p$. Set, $$x_n:=\sum_{j=1}^n\big|F(e_j)\big|^{q-1} \operatorname{sgn}\big(F(e_j)\big)e_j.$$ Here, as usal $e_j$ is the sequence having $j$-th term as $1$ and all ...

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This follows immediately from Holder's inequality. Take $q$ such that $\frac{1}{p}+\frac{1}{q}=1$. Then: $||X||_1=\int_{\Omega} |X|\leq (\int_{\Omega} |X|^p)^{\frac{1}{p}}(\int_{\Omega} 1^q)^{\frac{1}{q}}=||X||_p$ Edit: the general case follows from the special case. Assume $1\leq r<p<\infty$. We can define $Y=X^r$. By the special case we proved we ...

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You basically showed already that there does not exist a $q>1$ such that $X\subset L^q$ holds. Using the statement from the linked question, this implies that the condition $$\tag{*} X\subset \bigcup_{1<p\leq\infty} L^p(0,1)$$ is false (if it were true then $X\subset L^q$ would follow for some $q>1$). You might be falsely thinking that (*) should ...

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Here is a different solution based on the following fact: If $\frac1p+\frac1q=$, $f\in L_p(\lambda)$ and $g\in L_q(\lambda)$, then $f*g$ is uniformly continuous: Here is a short proof of this: $$|(f*g)(x+h)-(f*g)(x+k)|\leq \int|f(x+h-y)-f(x+k-y)||g(y)|\,dy\leq\|\tau_{-(k-h)}f-f\|_p\|g\|_q$$ The conclusion then follows by $L_p$ continuity of the translation ...

1

Begin a closed linear subspace of $L_1$, $X$ is itself a Banach space. Each $X_n$ is a closed subset of $X$: Suppose $f\in \overline{X}_n^X$ (the closure of $X_n$ relative to $X$). Then there is a sequence $\{f_k:k\in\mathbb{N}\}\subset X_n$ such that $\|f-f_k\|_{L_1}\xrightarrow{k\rightarrow0}$. By standard results (application of Chebyshev Markov, and ...

3

As far as I understand, you are able to prove that the operator $T_t: L^{\infty} \to L^{\infty}$ has norm $\|f\|_{L^1}$ if $f \in L^1$ is uniformly continuous. You can extend the assertion using a density argument. Let $f \in L^1$ be arbitrary, then there exists a sequence of uniformly continuous functions $(f_n)_{n \in \mathbb{N}} \subseteq L^1$ such that $... 3 Closedness of$X_n$'s follows immediately from Fatou's Lemma: If$\|f_j\|_{1+\frac 1 n} \leq n$for all$j$and$f_j \to f$in$X$then there is a subsequence$f_{j_i}$which converges a.e. so$\|f\|_{1+\frac 1 n} \leq \lim \inf_k \|f_{j_k}\|_{1+\frac 1 n} \leq n$. Suppose there is an open ball$B(f_0,r)$in$X$which is$ \subset X_{n_0}$. Let$f \in X$. ... 1 By the open mapping theorem (or its corollary), a bijective linear bounded map between Banach spaces has a bounded inverse. In this case, the identity mapping$I:(S,\|\cdot\|_\infty)\to (S,\|\cdot\|_2)$is clearly bijective, linear, and bounded by$\|f\|_2\le\|f\|_\infty$. The spaces are Banach as the asker rightly observes. Hence the inverse mapping (which ... 0 First of all note that the graph of$T$is precisely $$\mathscr G(T) : = \{(f,f)\ :\ f \in S \} \subseteq \left (S,\|\cdot\|_2 \right) \times \left (S, \|\cdot\|_{\infty} \right ).$$ To prove that$\mathscr G (T)$is closed we first take a convergent sequence$\{(f_n,f_n) \}_{n \geq 1}$in$\mathscr G (T)$converging to some$(f,g)$with respect to the ... 3 Pick$\alpha(x)=x$. Then we have$\Vert \alpha \Vert_\infty = 1$. Assume that there exists$f\in L^2$with$\Vert f \Vert_2=1$such that$1=\Vert M_\alpha \Vert = \Vert M_\alpha (f) \Vert_2$. For$0<\varepsilon <1$we have $$1^2=\Vert M_\alpha(f) \Vert_2^2 = \int_0^1 x^2 f(x)^2 dx = \int_0^{1-\varepsilon} x^2 f(x)^2 dx + \int_{1-\varepsilon}^1 x^2 ... 2 Yes you can. That can be achieve by direct application of Holder's inequality. For the specific case at hand, if f\in L_2(\mu)$$\|f\|_1=\int_X|f|\,d\mu=\int_X|f|\mathbb{1}\,d\mu\leq \|f\|_2\|\mathbb{1}\|_2=\|f\|_2(\mu(X))^{1/2}\leq(\mu(X))^{1/2}\varepsilon$$More generally, if 0<s<r and f\in L_r(\mu), then |f|^s\in L_{r/s}(\mu) and so$$ \... 6 $$\int|f|=\int|f|\cdot1\leq\bigg(\int|f|^2\bigg)^{1/2}\bigg(\int 1^2\bigg)^{1/2}=||f||_2\cdot \mu(X)^{1/2}.$$ More generally, for any$0<p<q<\infty$we have, $$f\in L^q\implies\int|f|^p=\int|f|^p\cdot1\leq \big|\big||f|^p\big|\big|_{q/p}\cdot ||1||_{q/(q-p)}=||f||^p_q\cdot \mu(X)^{(q-p)/q}$$$$\text{ That is }||f||_p\leq ||f||_q\cdot \mu(X)^{(1/p)-(... 0 Write C_0[-L/2,L/2] for the subspace of C[-L/2,L/2] consisting of the f with f(-L/2)=f(L/2). Then by Stone-Weierstrass, the f_n generate an L^2-dense subspace of C_0[-L/2,L/2] and we know C[-L/2,L/2] is L^2-dense in L^2[-L/2,L/2]. To complete the proof we need that C_0[-L/2,L/2] is L^2 dense in C[-L/2,L/2]. But C_0[-L/2,L/2] has ... 1 You salvage the approach by showing that the continuous functions such that f(-L/2)=f(L/2) are dense in L^2. This is easy to show as you can always consider a small interval around a point where you don't need to approximate your function. That is, given f\in L^2 and \varepsilon>0, find g continuous with \|f-g\|<\varepsilon/2. Then define ... 2 The proof is essentially correct. The fix is to prove the density instead on the torus, ie. [-L/2,L/2] with the endpoints quotiented together. Then they are the same point, so there’s nothing to separate! (Note that S-W still works on a compact Hausdorff space.) Then since null sets don’t matter, L^2 on the torus is naturally identified with L^2 on ... 1 Answer of (2) In this answer, I will use L^2,L^\infty for simplicity instead of L^2\big([0,1],\Bbb K\big) and L^\infty\big([0,1],\Bbb K\big). Also, these are sets of equivalence classes, but I will not distinguish in between equivalence classes and their representatives, as any two representatives are almost everywhere equals. From the above, we have ... 2 For p\in (1,\infty]. Let C be the set of all x=(x_n)_{n\in \Bbb N} in l^p such that f(x)=\sum_{n\in \Bbb N}x_n converges. Then$$ (\bullet)\quad \sup_{0\ne x\in C}|f(x)|/\|x\|=\infty.$$Let C_0=\{x\in C: f(x)=0\}. Then C_0 is dense in C. Proof: If p\in C and f(p)\ne 0 then for any \epsilon \in \Bbb R^+ there exists q\in C with \|q\|&... 3 The result is true only for p\in(1,\infty]. Upper index n will denote the nth sequence; lower index k will denote the kth term of the sequence. Consider first an auxillary sequence$$x^n = \left(1,-1,\frac12,\frac{-1}2,\dots,\frac1n,\frac{-1}n,0,0,\dots\right)$$The termwise and \ell^p limit$$x= \left(1,-1,\frac12,\frac{-1}2,\dots,\frac1n,\frac{-... 0 Actually, the property you are pointing out is fairly more general. In fact, for any$1\leq p < q \leq +\infty$, if you have a function $$f\in L^p(\mathbb{R})\cap L^q(\mathbb{R}),$$ then,$f$belongs to all$L^r(\mathbb{R})$for all$r\in(p,q)$. In fact, let assume that$q<+\infty$(since the proof for$q=\infty$is already in the other answers). Let'... 2 For$f\in L^1$we have$\int|f|<\infty$. $$f\in L^\infty\cap L^1\implies \int|f|^2=\int|f|\cdot |f|\leq ||f||_\infty\int|f|<\infty\implies f\in L^2.$$ Note that for any measurable$f$we define$||f||_\infty=\inf\big\{M:\lambda(x:|f(x)|>M)=0\big\}$. Now, from definition of infimum we have$\lambda\big(\{x:|f(x)|>||f||_\infty+1/n\}\big)=0$for ... 1 It is not true. Short version: you can join a uniformly convergent and a pointwise but not uniformly convergent sequence with the same limit in one sequence and generate a counterexample. Now let us explicitly write down a counterexample. We choose $$f_n(x)= \sqrt{x^2+\frac 1n} \quad\mbox{for even }n\qquad\mbox{and}\quad f_n(x)=|x|+\frac xn \quad\mbox{for ... 3 If x \in \ell^p, then$$ \sum_{n=1}^{\infty}|x_n|^p<+\infty.$$In particular,$$\lim_{n}|x_n|=0.$$The last equality implies that there is a n_0 such that |x_n| \le 1 for every n\ge n_0. Therefore, for every n \ge n_0 and p'>p, |x_n|^{p'} < |x_n|^p. Then, by the comparison test:$$ \sum_{n=1}^{\infty}|x_n|^{p'}<+\infty$$and x \in \... 2 Since L^p(\mu) is reflexive, its unit ball B is weakly compact. Since T is continuous for the norm topologies, it is also continuous for the weak topologies and hence T(B) is a weakly compact set in C(X). In particular, since the map f \mapsto f(x) is a continuous linear functional on C(X) for each x \in X, for any sequence f_n in T(B) ... 2 \|f_n-f\|_1 \to 0 implies there is a subsequnce f_{n_k} converging a.e. to f. So \int |f|^{p} \leq \lim \inf \int |f_{n_k}|^{p} \leq 1. 0 If f_n \to f in L^1 then f_n \to f in measure. Then use this Exercise on convergence in measure (Folland, Real Analysis) 0 V^\perp consists of all functions written in terms of the inner product \langle f,g\rangle=\int_0^{\pi}f(t)g(t)dt:$$ f-\frac{\langle f,\sin(x)\rangle}{\langle \sin(x),\sin(x)\rangle}\sin(x)-\frac{\langle f,\cos(x)\rangle}{\langle\cos(x),\cos(x)\rangle}\cos(x),\;\;\; f\in L^2[0,\pi]. $$This is because \langle \sin(x),\cos(x)\rangle =0. 1 Hint: Use Young's inequality for each summand. This will look similar to your assumed inequality (maybe you meant to add the exponents so that it becomes Young's inequality). edit: To address the edit of the question: Yes, there does seem to be a mistake. In the second part the exponents p and q are missing over the parts in the parentheses. 1 Here is what is wrong: the correct statement is that the separability of L^p, 1\le p <\infty, is equivalent to the existence of a countable collection C of measurable sets of finite measure such that for each measurable set Q of finite measure and for each positive r there exists a set A\in C such that the measure of Q\Delta A is less that ... 0 An elegant way to prove the inequality \|\int f d\mu\|\le \int \|f\| d\mu for functions with values in a Banach space (X,\|\cdot\|) is the norm formula \|x\|=\sup\{ |\phi(x)|: \phi\in X^*, \|\phi\|^*\le 1\} (which is a consequence of Hahn-Banach). For \phi\in X^* with \|\phi\|^*\le 1 you get$$ \left|\phi(\int fd\mu)\right|=\left|\int \phi(f)d\mu\... 1$\Bbb T$is the circle parametrized by$e^{2\pi ix}$with$x\in[0,2\pi]$. So you can view any function defined on$\Bbb T$as a function on the interval$I=[0,2\pi]$and $$\int_{\Bbb T}f=\int_If(x)dx\;.$$ Now take$f\in L^q$, that is, the integral$\int_I|f|^q$is finite. Partition$I$into subset$A$where$|f(x)|\le 1$and subset$B$where$|f(x)|>1$. ... 1 The space$L_p (X)$is dense subspace of$L_1 (X)$and$S$is defined on$L_p (X).$So by Hahn - Banach Theorem you can extend$S$to an operator$S_1 $on whole$L_1 (X)$but since$L_p (X)$is dense in$L_1 (X)$the extension must be unique, since every$f\in L_1 (X) $can be approximate by functions$f_n\in L_p (X) $such$||f_n -f||_1 \to 0.$-1 The correct statement is that the space of summable functions (modulo a.e. equivalence) is separable if and only if there is a countable collection C of measurable sets of finite measure such that for each measurable set Q of FINITE measure and each integer N there is a set A in the collection C such that the measure of the symmetric difference between Q and ... 1 For$L^p$-spaces, you're reasoning is correct: We may extend the functions by$0$and get a continuous embedding. In Sobolev Spaces, the same technique works if the Sobolev functions vanish on the boundary of the smaller domain$A$: $$W^{k,p}_{0}(A) \hookrightarrow W^{k,p}_{0}(B)$$ You can easily check that the weak differentiability is not affected. If we ... 2 If one consider real valued functions, i.e.$L_2(X;\mathbb{R})$with$\langle f,g\rangle =\int fg\,d\mu$Then $$F(f+h)=\langle f+h,f+h\rangle = \langle f,f\rangle + 2\langle f,h\rangle +\langle h,h\rangle=F(f)+2\langle f,h\rangle + F(h)$$ Cearly,$h\mapsto \langle f,h\rangle$is linear, and$\frac{F(h)}{\|h\|_2}\xrightarrow{\|h\|_2\rightarrow0}0$. From this,... 1 Let$u_{n}(x)$be a function of piecewise linear bumps of height$\frac{1}{n}$and width$\frac{1}{n}$on$\Omega=(0,1)$. The weak derivative is piecewise constant and satisfies almost everywhere$|u_{n}'(x)|=1$. Thus$||u_{n}'||_{L^q} = 1$. Then again,$||u||_{L^p}$vanishes as$n$goes to infinity. So you're inequality can't exist. 3 You have$g := |\varphi|^2 \in L^1(\mathbb{T})$and $$\widehat{g}(n) = \int_{\mathbb{T}} g(x) \cdot \overline{x}^n \, dx = \int_{\mathbb{T}} e_{-n}(x) \cdot |\varphi(x)|^2 \, dx = 0$$ for any$n \in \mathbb{Z}$with$n \neq 0$. Furthermore $$\widehat{g}(0) = \int_{\mathbb{T}} |\varphi(x)|^2 \, dx = 1.$$ Now check that the constant function$1 \in L^1(\mathbb{...

0

Write $h(p)=\int f^pd\mu$. $h'(p)=\int f^p\log(f)d\mu.$ Then we have $$\frac{d}{dp}(h(p)^{1/p})=\frac{d}{dp}(e^{\frac{\log(h(p))}{p}})=e^{\frac{\log(h(p))}{p}} \frac{\frac{p}{h(p)}h'(p)-\log(h(p))}{p^2}$$ When p=1, the above evaluates to $$h'(1)=\int f^p \log(f)d\mu.$$

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The map $$F: (L^\infty(\mathbb{T}, |f \, d\lambda|), \|\cdot\|_1) \to \mathbb{C}, \, g \mapsto \int_{\mathbb{T}} fg \, d\lambda$$ is a bounded linear functional and hence continuous since $$|F(g)| = \left|\int_{\mathbb{T}} fg \, d\lambda \right| = \left|\int_{\mathbb{T}} g \, d(f\, d\lambda)\right| \leq \int_{\mathbb{T}} |g| \, d |f \, d\lambda| = \|g\|_1$$ ...

3

Just to write it out: So you know that $X^{(m)}$ converges to $X$ in $L^2([0,T]\times \Omega)$ for every $T>0$. Hence, let $X^{(m,1)}$ be a subsequence of $X^{(m)}$ converging a.e. on $[0,1]\times \Omega$ and recursively, let $X^{(m,n+1)}$ be a subsequence of $X^{(m,n)}$ which converges to $X$ a.e. on $[0,n+1]\times \Omega$, which, of course, we can do ...

0

Let $\mu_Y:=\mathsf{E}Y$, $\mu_X:=\mathsf{E}X$, $\Sigma_X:=\operatorname{Var}(X)$ $\Sigma_Y:=\operatorname{Var}(Y)$, and $\Sigma_{X,Y}=\mathsf{E}X(Y-\mu_Y)$. Consider the following transformation (assuming that $\Sigma_X$ is invertible): $$Z:=Y-\Sigma_{X,Y}^{\top}\Sigma_X^{-1}X.$$ Since \begin{align} \mathsf{E}X(Z-\mathsf{E}Z)^{\top}&=\mathsf{E}X(Y-\...

0

Suppose $f_{X,Y}$ is the joint density of the variables with $Y$ included. This is a Gaussian density function by assumption. The conditional mean is almost surely equal to the mean of the conditional density $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{\int_{\mathbb{R}}f_{X,Y}(x,y)dy}.$$ Namely, you can show that $$E(Y|X) = \int_{\mathbb{R}} yf_{Y|X}(y|X)dy \;\;... 6 \frac{1}{\log (2+n)}\in c_0. But since for x\gg 1, \log x\le C_q x^q for any q> 0, in particular (\log x)^p \le C_p x for all p\in[1,\infty), so$$\frac1{(\log (2+n))^p} \ge \frac{C_p}{2+n} which is not summable.

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