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13

The area is known, $$\frac{4 \Gamma \left( 1 + \frac{1}{p} \right)^2}{\Gamma \left( 1 + \frac{2}{p} \right)}$$ The method is due to Dirichlet (1839) but they appear to have blocked out the relevant pages in that article. Alright, found it ELSEWHERE, page 389. A little hard to read; that is the sort of thing that happens if you scan something and use ...

8

Using Will Jagy's formula and a numerical solver, $p \approx 2.10134909469$ does the trick to within the accuracy of my calculator. Incidentally, $p \approx 2.00208615381$ gives $\pi = 22/7$. And $p \approx 1.79147384986$ gives $\pi = 3$, compliant with 1 Kings 7:23...

4

Consider a sequence $x_n \in C[-1,1]$ such that a) $x_n(t) = 1$ for all $t\in [-1, -1/n]\cup [1/n,1]$ b) $x_n(0) = -1$ c) $x_n(t)$ is the line joining $(-1/n,1)$ and $(0,-1)$ on $[-1/n,0]$ d) $x_n(t)$ is the line joining $(0,-1)$ and $(1/n,1)$ on $[0,1/n]$ Then, $\|x_n\| = 1$ for all $n$, and $$f(x_n) \sim 2 - \frac{2}{n} + 2 \to 4$$ Hence, $\|f\| = ... 2 Do it first for the case$\ln|f|$is bounded. Use$|f|^p = \exp(p\,\ln|f|) = 1 + p \ln|f| + O(p^2)$as$p \to 0$. Then $$\left(\int |f|^p \, d\mu\right)^{1/p} = \left(1 + p\int \ln|f| \, d\mu + O(p)^2\right)^{1/p}$$ and use$(1+px + O(p^2))^{1/p} \to e^x$as$p\to 0$. To get it for unbounded functions, you probably have to use some kind of dominated or ... 2 The proof can be found in Section 6.8 of the old book Inequalities by Hardy, Littlewood and Polya. As Stephen Montgomery-Smith remarks in his answer, the proof is rather easy if$\log |f|$stays bounded. In the general case, Hardy suggests to write $$\int \log |f| \leq \log \|f\|_q \leq \int \frac{|f|^q-1}{q}$$ and to remark that, for$t>0$,$q \mapsto ...

2

I'll go through the proof for a couple of the axioms for a vector space. $C(X)$ has a zero element Let $0\colon X\to\mathbb{R}$ be given by $0(x)=0$ for all $x\in X$. We see that for any $f\colon X\to\mathbb{R}$ we get $(0+f)(x)=0(x)+f(x)=0+f(x)=f(x)=(f+0)(x)$ and so $f+0=f=0+f$, hence $0$ acts as a zero element of $C(X)$. The addition in $C(X)$ is ...

2

The case $n=1$ holds always. For $n\geq2$, i don't have any kind of general answer, but I expect such inequality to hold basically for norms that behave like the one-norm, and little else. As an example, the inequality fails for any norm in the $2\times2$ matrices. Indeed, if $$A=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ ... 2 Yes, this is true. Using Holder inequlity you can show that \Vert f\Vert\leq\Vert a\Vert_q. Now consider$$ x=(|a_1|^{q-1}\operatorname{sign}(a_1),\ldots,|a_n|^{q-1}\operatorname{sign}(a_n)) $$then f(x)=\Vert a\Vert_q^q and \Vert x\Vert_p=\Vert a\Vert_{q-1}^q. So \Vert f\Vert\geq |f(x)|/\Vert x\Vert_p=\Vert a\Vert_q. From these inequalities it ... 2 Let's break it down like this: Given a metric space (Y,d), a set X, and a bijection f\colon X \to Y, the pull-back of d via f, f^\ast d \colon X\times X \to [0,\infty);\; f^\ast d(a,b) = d(f(a),f(b)) is a metric on X. Thus f\colon (X,f^\ast d) \to (Y,d) is an isometry, in particular a homeomorphism. If, in the situation above, X carries a ... 1 When you take limits, inequalities become weak. So the assumption that ||s_n-t||<r for all n only allows you to conclude that \lim_{n\to\infty}||s_n-t||\leq r. (Think about 1-\frac{1}{n}\to 1 as n\to\infty; every term is less than 1, but the limit is not). (As an aside, you have some norms missing in your proof that ... 1 For the second inequality I refer you to Brezis Corollary 9.14. For the first inequality, remember that \|u\|_\infty\leq \|u\|_{C^{0,\gamma}(\overline{U})}, therefore$$\int_U |u|^p\leq\int_U\|u\|_\infty^p\leq |U|\|u\|_{C^{0,\gamma}(\overline{U})}^p$$where |U| is the Lebesgue measure of U. 1 The first inequality holds as mentioned by Tomas. The second inequality I think is a direct result of a Theorem which extends from Morrey's Inequality: If U is bounded, open subset of \mathbb{R}^{n}, and suppose \partial U is C^{1}. If n < p \leq \infty and u \in W^{1,p}(U). Then u has a version u^{*} \in C^{0,\gamma}(\bar{U}) such that ... 1 For p\ge2 and u\ge0, Jensen's inequality gives$$ \begin{align} \left(\frac{1+u^2}{2}\right)^{p/2}&\le\frac{1+u^p}{2}\tag{1}\\ \left(1+u^2\right)^{p/2}&\le2^{p/2-1}\left(1+u^p\right)\tag{2} \end{align} $$Thus,$$ \begin{align} \frac{u^{p-2}(1+u^2)}{1+u^p} &=\left(\frac{u^p}{1+u^p}\right)^{1-2/p} ...

1

Continue the proof of Davide Giruado. As $B(x_0,r) \subset E$ then $B(0,r)=B(x_0,r)-x_0 \subset E$ Let $x$ be the arbitrary element of $X$, Choose $n$ such that $\left \| \frac{x}{n} \right \|< r$. So $\frac{x}{n} \in E$ which gives $x\in E$. Consequently $X\subset E$ which deduces that $X=E$ as claimed.

1

In the one direction, to construct a compact set with dense span from a countable dense subset, note that if $(x_n)_{n\in\mathbb{N}}$ is a sequence converging to $x_\ast$, then the set $\{ x_n : n \in \mathbb{N}\} \cup \{x_\ast\}$ is compact. Construct a convergent sequence from the countable dense subset without changing the span. In the other direction, ...

1

I have already posted the answer here where somebody wanted to know for which $p$ is $\pi=42$. Using $$\pi_p=\frac{2}{p}\int_0^1 [u^{1-p}+(1-u)^{1-p}]^{1/p}du$$ we get $p=2.60513$ and $p=1.623$ where $\pi_p=3.2$ and we also know that those are the only two answers. Reference: http://www.jstor.org/stable/2687579

1

I don't understand why your polynomials do not include a constant term $a_0$. I included it below. Let's first recognize that bounding $P(x_0)$ by $M_n\|P\|$ where "$M_n$ is a constant that depends on the degree of $P$" is not specific enough for conclusion either way. Just saying "depends on the degree" does not mean it cannot be bounded independently ...

1

That's not even a norm. Take $a = 0$, $b=1$, $x(t) = -2 + 2t$. Did you mean "$|x(a)|$"? By the way, "\max" gives you $\max$ and is easier on the eyes. EDIT: after I wrote the above, the question was fixed and "$x(a)$" was replaced by "$|x(a)|$", so the given norm (I'll call it "$\| \cdot \|_l$") actually is a norm. The norm $\| \cdot \|_l$ is equivalent ...

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