# Tag Info

4

For $t > 0$, write $$t^\theta = t^\theta\cdot 1^{1-\theta}.$$ Taking the logarithm of both sides, what you need to show becomes $$\theta\log t + (1-\theta) \log 1 \leqslant \log (\theta\cdot t + (1-\theta)\cdot 1),$$ which follows from the concavity of $\log$.

4

Since $$\lim_{h\to0}\int_0^x\left|\frac{f(t+h)-f(t)}{h}-g(t)\right|^2\,\mathrm{d}t=0\tag{1}$$ Hölder's Inequality says $$\lim_{h\to0}\int_0^x\left|\frac{f(t+h)-f(t)}{h}-g(t)\right|\,\mathrm{d}t=0\tag{2}$$ There is no way to derive the continuity of $f$ since changing $f$ on a set of measure $0$ will not affect $(1)$. However, the Lebesgue Differentiation ...

4

Homeomorphisms, in general, preserve neither completeness nor total boundedness. Consider $\Bbb R$ and $(0,1)$.

3

Your second idea works fine: just observe that the function $x\to f_1(x)f_2(x)$ is $L^1$ (you proved that with Hölder's indequality).

3

Yes, it's complete. First, observe that your norm is comparable to the simpler norm $$\|f\| = \sup_{n\in\mathbb{Z}} \int_n^{n+1}|f(y)|\,dy \tag{1}$$ Indeed, $\|f\|\le |f|\le 2\|f\|$ because every interval of length $1$ is contained in some interval of the form $[n,n+2]$. The space with $(1)$ is just the direct sum $\bigoplus_\infty X_n$ of Banach spaces ...

3

To expand a bit on the standard answer ("$p=2$ is a Hilbert space and Hilbert spaces are nice"), let's talk about why Hilbert spaces are nice. One of the key feature of Hilbert spaces is that they are self dual. This is expressed through the Riesz representation theorem, which says that if I want to define a bounded, linear map $\phi:L^2\rightarrow ... 3 I elaborate on a remark in one of the other answers. Consider a measure space$(X,\mathcal{M},\mu)$which has sets of arbitrarily small measure and sets of arbitrarily large measure. (Here I think of the Lebesgue measure on the real line.) Let$1 \leq q < \infty$and choose $$g \in L^q \setminus \bigcup_{p \in [1,\infty), p \neq q} L^p.$$ This is ... 2 I think part of it is because of Holder's inequality, which dictates that $$|fg|_{1}\le |f|_{p}|g|_{q}$$ But the other reasons goes into this is much deeper. For example, this duality is "structural" as it holds regardless of the measure space is$\sigma$-finite. And this has to do with concepts like uniform convexity. I used to wondering the same thing ... 2 In the case$q<p^*$the Rellich–Kondrachov theorem applies, if your domain$\Omega$has the required properties. If$q=p^*$, the$L^q$norm is still controlled by the$W^{1,p}$norm; and since a weakly convergent sequence is norm-bounded,$L^q$norm is bounded. Moreover, you will have weak convergence in$L^q$, since linear maps preserve weak ... 2 Note that this is a special case of the general theorem of means. If$a, b$are non-negative and$0 < x, y < 1$with$x + y = 1$then$a^{x}b^{y} \leq ax + by$. Equality occurs if$a = b$. Here in the current question$a = t, b = 1, x = \theta, y = 1 - \theta$. An easy proof of the general theorem mentioned above is based on Mean Value Theorem. ... 2 If we take$g:\theta\to t^{\theta}$we are just stating that the graphics of$g$on the interval$(0,1)$lies below the line through$(0,1)$and$(1,g(1))$. However, that is trivial, since$g(\theta)$is a convex function due to: $$g''(\theta) = t^{\theta}\log^2 t \geq 0.$$ 2 It is certainly not true that$|f|^2 < |f|^4$(think of the case where$|f|<1$). Hint: Cauchy-Schwarz on$\int |f|^2 = \int g h$where$g = |f|^2$and$h = 1$. 2 Write$|f|^2 = |f|^2 \cdot 1$and apply the Cauchy-Schwarz inequality to get$\||f|^2\|_1 \le \||f|^2\|_2 \|1\|_2$. Simplify the inequality. Note if$f \in L_4([0,1])$, then the inequality$\|f\|_2 \le \|f\|_4$implies$\|f\|_2 < \infty$and hence$f \in L_2([0,1])$. 2 Hints: Fix$\epsilon>0$. Fix$k \in \mathbb{N}$. Using Markov's inequality, show that $$A_N^k := \left\{x; \exists n \geq N: |f_n(x)-f(x)| \geq \frac{1}{k} \right\}$$ satisfies $$m(A_N^k) \leq k \sum_{n=N}^{\infty} \|f_n-f\|_{L^1}.$$ Conclude from the first step that there exists$N=N(k)$such that$m(A_{N(k)}^k) \leq \epsilon 2^{-k}$. Set $$A := ... 2 This is not true. In fact, Kolmogorov constructed (1923) an example of a L^1 function whose Fourier series diverges almost everywhere (later improved to everywhere divergent). Om the other hand, if f\in L^p for some p>1, it's a deep theorem by Carleson (the p=2 case) and Hunt (p>1) that the Fourier series of f converges pointwise almost ... 2 Consider \mathbf{x} = (1,0,0,0,\ldots) and \mathbf{y} = (0,1,0,0,\ldots). Then \mathbf{x+y} = (1,1,0,0,\ldots) and \mathbf{x-y} = (1,-1,0,0,\ldots).$$\|\mathbf{x}+\mathbf{y}\|_p^2 + \|\mathbf{x}-\mathbf{y}\|_p^2 = 2\times(1+1)^\frac{2}{p} = 2^{1+\frac{2}{p}}$$Meanwhile,$$2(\|\mathbf{x}\|_p^2+\|\mathbf{y}\|_p^2) = 2\times(1+1) = 4$$If the ... 1 From Minkowski, \| \sum_1^n |f_k|\|_p \le \sum_1^n \|f_k\|_p \le \sum_1^\infty \|f_k\|_p. Now raise to the pth power. 1 Triangle inequality then dominated convergence theorem. Edit: By triangle inequality, we have,$$\int G^p= \lim \int G_k^p=\lim \int (\sum_{k=1}^n |f_k|)^p=\lim ||(\sum_{k=1}^n |f_k|) ||^2\leq \lim \sum_{k=1}^n ||f_k||^2$$By DCT, we then conclude the given inequality. 1 Another proof. More elementary than Urban's proof. (But less general.) Consider the subspace c \subseteq l^\infty consisting of the real sequences that converge. So c is a closed linear subspace of l^\infty. Let L \;:\; c \to \mathbb R be the "limit" linear functional. That is, for x = (x_1,x_2,\cdots), define$$ L(x) := \lim_{n\to\infty} ... 1 This is not true because of the following Theorem$\textbf{Theorem:}$If$X$is a normed space such that its dual$X'$is separable, then$X$itself is separable. So, if$(l^{\infty})' =l_1$, then it will follow that$l^{\infty}$is separable which is not true. 1 In general, a conformal map preserves angles (but not necessarily length). In the case of an inner product space, since the angle between two vectors$v,w$is defined to be the arccosine of thrle quantity$\frac{\langle v,w\rangle}{|v||w|}$, a conformal map$c:V\to W$is one that preserves this quantity. In particular, since$|v| = \sqrt{\langle ...

1

Follow your idea, $s = \|x\|$, Consider $A = \{s < 2\}$, $B = \{s > 2\}$. On $A$, we have $$f\le \dfrac{C}{s}$$ for some constant $C$. because $$\inf_{0 \le s < 2} s^{1/2}\log s$$ is bounded. $$\int_A |f|^p\le \int_{A} \dfrac{C}{s^p} dxdy = \int_{0}^{2\pi}\int_{0}^{2}\dfrac{C}{s^p}s ds d\theta = 2\pi\int_0^{2}\frac{C}{s^{p-1}}ds$$ it is ...

1

This is a broad question with no definitive answer. You are asking for a description of characteristic functions $\chi_S$ that are Fourier multipliers bounded on $L^p$, or $C^\alpha$, etc. Certainly, this is so when $S$ finite or $\mathbb{Z}\setminus S$ is finite. The best known nontrivial case is $S=\{n\in \mathbb{Z}:n\ge 0\}$ which is directly related to ...

1

It suffices to show that $$\sum_{k=0}^\infty \int_{A_{2^k}} |f|<\infty \tag{1}$$ because $\bigcup_{k=0}^\infty A_{2^k}$ covers the plane except for a subset of finite measure (and on a subset of finite measure $L^{3/2}\subset L^1$). Naturally, Hölder's inequality comes into play: $$\int_{A_{2^k}} |f| \le \left(\int_{A_{2^k}} |f|^{3/2}\right)^{2/3} ... 1 If such a function exists then$$\tag{1} f(t)=\frac1\pi\sum_{n=1}^\infty B_n\sin(nt). $$But then we would have: \begin{eqnarray} \infty>\|f\|^2&=&\langle f,f\rangle=\frac{1}{\pi^2}\sum_{m,n}B_mB_n\int_{-\pi}^\pi\sin(mt)\sin(nt)\,dt=\frac{1}{\pi^2}\sum_{m,n}\pi B_mB_n\delta_{m,n}\\ &=&\frac1\pi\sum_{n=1}^\infty ... 1 As Farnight suggested, it is better to use \sup. I also prefer to work on a more concrete level: with numbers, not with sets of numbers. So I would write: "for n\in \mathbb{N} ,$$|v(n)|=\left|\frac{n\sin{(n!)}}{n^2+1}\right|\le \frac{n }{n^2+1}\le \frac{n^2}{n^2+1} <1 $$Therefore, the sequence v is bounded." Note that your write up leaves the ... 1 You don't need Zorn's lemma. The function f can be defined explicitly as$$ f_1\chi_{\{|x|\le 1\}}+\sum_{n=2}^\infty f_{n} \chi_{\{n-1<|x|\le n\}} $$which is evidently in L^1_{\rm loc} (on every bounded interval, only finitely many terms are nonzero). The convergence of f_n to f on every bounded interval follows from the fact that the restriction ... 1 When \Gamma is uncountable, c_0(\Gamma) is not separable. It consists of all functions g : \Gamma \to \mathbb R (with countable support) such that for every \epsilon>0, the set \{\gamma \in \Gamma : |g(\gamma)|>\epsilon\} is finite. But (you may ask) even if this particular pre-dual is nonseparable, might there not be some other pre-dual ... 1 We have$$ c_0(\Gamma) := \{x \in \mathbf K^\Gamma \mid \forall \epsilon > 0 \,\exists \Gamma' \subseteq \Gamma, |\Gamma'| < \infty \land \sup_{\gamma \in \Gamma - \Gamma'} |x(\gamma)| < \epsilon \} $$which is a Banach space with respect to the norm$$ \def\norm#1{\left\|#1\right\|}\norm{x}_\infty := \sup_{\gamma \in \Gamma} ...

1

The argument is the following: Let $v_n$ be a sequence in $W^{1, p}_0(\Omega)$ for some $\Omega \subset \mathbb R^N$ and $p >N$. If $v_n \to v$ weakly in $W^{1,p}_0(\Omega)$, then $||v_n||_{1, p}$ is uniformly bounded. By the Sobolev Embedding (Theorem 7.17) and the fact that $$C^{0, \alpha}(\overline \Omega) \to C(\overline \Omega)$$ is compact, there ...

Only top voted, non community-wiki answers of a minimum length are eligible