Tag Info

Hot answers tagged

6

You have $$ \phi(p) = \int_0^1 \lvert f(x) \rvert^p \, dx $$ $p \mapsto x^p = e^{p\log{x}}$ is smooth for $x \geqslant 0$, so we can differentiate twice with respect to $p$, $$ \phi''(p) = \int_0^1 \lvert f(x) \rvert^p (\log{\lvert f(x) \rvert})^2 \, dx, $$ which is obviously nonnegative. (This is consistent even for $f$ possessing zeros if we make the ...


6

Actually your question is the combination of two questions. About your first question: You certainly know the Euclidean norm, $$\left\|v\right\|_2 = \sqrt{\sum_{k=1}^n \left|v_k\right|^2} = \left(\sum_{k=1}^n \left|v_k\right|^2\right)^{\frac{1}{2}}$$ derived from the Euclidean scalar product. Note that I used absolute value bars here, which are superfluous ...


5

Let's just think about norms in the plane. The ordinary Euclidean norm is $(x^2 + y^2)^{1/2}$. The unit circle for that norm is the ordinary unit circle. There are other useful norms. One is $|x| + |y|$. Its "unit circle" is the square with vertices $(\pm 1, 0)$ and $(0, \pm 1)$. These norms are part of a family of norms defined by $\|(x,y)\|_p = (|x|^p + ...


5

Hint: use the Holder inequality.


4

Let $p,q$ such as $1\leq p<q$ and $x = \{x_k\} \in l^p$. Then by definition $\sum_{k=1}^{\infty} {|x_k|^p} < \infty$. Therefore $|x_k| \to 0$ as $p \to \infty$, hence $|x_k| \to 0$. This implies that there exists an index $k_0$ such that $|x_k| < 1$ for any $k ≥ k_0$. So for $k ≥ k_0$ we have $|x_k|^q < |x_k|^p$ which implies ...


4

Suppose $f$ is unbounded. Then there exists a sequence $\{x_n\}$ with $x_n\to0$ and $f(x_n)=1$ for all $n$. Now choose any $y\in X$. Then $y-f(y)x_n$ is in $\ker f$, since $f(y-f(y)x_n)=f(y)-f(y)=0$. Thus $y=\lim_n y-f(y)x_n$, where the elements in the sequence $\{y-f(y)x_n\}$ are in $\ker f$. So $\ker f$ is dense in $X$.


4

How about this: define $T : \ell^1 \to \ell^1$ by $$T(a_1, a_2, a_3, \cdots, a_n, \cdots) = (0, (1-1/2)a_2, (1-1/3)a_3, \cdots, (1-1/n)a_n, \cdots)$$ Then $$||T|| = \sup_{||x|| = 1} ||Tx|| = \sup_n (1-1/n) = 1$$ However, for all non-zero $x \in \ell^1$, $||Tx|| < ||x|| = ||T|| \cdot ||x||$.


4

If $X$ is only topological, "bounded" has no meaning. If $X$ is a metric space, the answer is "yes", as the closure consists of points that are within $\epsilon$ of points in the set.


3

No, for the same reason that the result doesn't work for functions (just define an antiderivative, and so on). The result for functions is discussed in this question. Similarly for series and sequences, basically by using the integral test.


3

You can formalize this: Take two Hilbert spaces $H\subset V$ such that The inclusion is dense (i.e. the image of $H$ in $V$ is dense in the topology of $V$) and, The inclusion is continuous (i.e. $H$ has a stronger norm than $V$). Then, if we identify $V$ with its dual (let's assume the spaces are real, though you can do this in general), we have ...


3

There are two versions of your question: The isometric one: is $C[0,1]$ isometric to a dual Banach space? The isomorphic one: is $C[0,1]$ isomorphic to a dual Banach space? Of course, the negative answer to 2. implies the negative answer to 1. However 1. is elementary. Morally, the answer is no because $C[0,1]$ has too few extreme points. To be more ...


3

$C(X)$ is defined to be the set of continuous functions $X\rightarrow \mathbb{R}$. To say that something is a function from $X$ to $Y$ requires that it is defined at every element of $X$. So saying that $f$ is a vector in $C[0,3]$ or $C[-3,0]$ is saying that it is a continuous map defined on $[0,3]$ and $[-3,0]$, respectively. However, given a vector ...


3

Any prime ideal is primary. $(0)$ is prime in any integral domain. There are integral domains with more than one maximal ideal.


3

I don't know if the assertion is true. What one can prove is that $p,q$ are Murray -von-Neumann equivalent if they fulfill $$ \|aa^*-p\|<\varepsilon,\quad \|a^*a-q\|<\varepsilon, $$ for sufficiently small $\varepsilon$, if the $C^*$-algebra $A$ is unital. Maybe one can streamline the proof to obtain it for $\varepsilon=1/4$. First note that ...


3

Consider $g(u) = \frac12 f(u^{-1/2}) u^{-3/2}$. Then for each $n$: $$ 0 = \int_a^1 t^{-2n} f(t) dt = \int_1^{a^{-2}} u^n g(u) du $$ so $g = 0$ (a classical application of the Stone Weiertrass theorem) and $f = 0$.


3

$T$ is bounded for all $\beta > -d$. The strategy is to show that $T$ is bounded both as a function $L^1(d\mu) \rightarrow L^1(d\mu)$ and as a function $L^\infty \rightarrow L^\infty$; the Riesz-Thorin theorem then shows that it is bounded $L^p(d\mu) \rightarrow L^p(d\mu)$ for any $p$. We just need the following estimate; everything else is standard. ...


3

Suppose $X$ is not bounded. Fix $x\in X.$ Claim: The collection of $d(x,y),~y\in X$ is not bounded. If it were bounded by some $C,$ then for any $y,z\in X$ we would have \begin{equation*} d(y,z)\leq d(y,x)+d(x,z)\leq 2C \end{equation*} which is a contradiction. Since the definition of sequential compactness needs every infinite sequence to have a ...


2

Your examples are lacking squares (think of homogeneity), but apart from that they are both valid norms for $H^4(I) \cap H^2_0(I)$, as is the normal $L^2$-norm. What you really want in most situations is not just any norm, but a norm with which the space is complete. This is not true for the $H^2$-norm: There holds by definition $$C^\infty_0(I) \subset ...


2

Remarks: The argument below proves the following. Let $1<r<\infty$ and $f\in L^r(\Omega, \Sigma, \mu)$ for some probability space $(\Omega, \Sigma, \mu)$. Then, the function $\phi:[1,r]\to$$\mathbb R$ defined by $\phi(p)=\|f\|^{p}_{L^p}$ is convex. Choose $\theta\in[0,1]$ and $p,q\in[1,r]$. Then, $\theta p+(1-\theta)q\in[1,r]$. Note that, by the ...


2

The multiplication operator $(Mf)(x)=xf(x)$ on $L^{2}[0,1]$ is a classical example of an operator with no eigenvalues, but its spectrum is $[0,1]$. $M$ has no eigenvalue because $Mf=\lambda f$ gives $(x-\lambda)f=0$, which forces $f(x)=0$ a.e.. To see that $[0,1]\in\sigma(M)$, note that the constant function $1$ is not in the range of $(M-\lambda I)$ for ...


2

Spectral theory in infinite-dimensional spaces is quite a bit more complicated than in the finite-dimensional case. In particular, we have to distinguish between the spectrum $\sigma(A)$ of an operator and its eigenvalues. Let $A$ be a linear operator on a Banach space $X$ over the scalar field $C$. We have $$ \sigma(A) = \{ \lambda \in C: (\lambda I - A) ...


2

In what follows, we assume that $ G $ is a locally compact Hausdorff group that does not have to be abelian. You can certainly define an involution $ ^{*} $ on $ {L^{1}}(G) $ by $$ \forall f \in {L^{1}}(G), ~ \forall x \in G: \quad {f^{*}}(x) \stackrel{\text{df}}{=} \overline{f(x^{-1})} \cdot \Delta(x^{-1}), $$ where $ \Delta $ denotes the modular ...


2

It's a simple calculation error. When substituting to convert $\int_a^b \lvert f(x)\rvert^2\,dx$ into the integral over $\lvert g(t)\rvert^2$, we have $$x = \frac{b-a}{2\pi}t + \frac{a+b}{2},$$ and hence obtain $$dx = \frac{b-a}{2\pi}\,dt.$$ So $$\int_a^b\lvert f(x)\rvert^2\,dx = \int_{-\pi}^\pi\left\lvert ...


2

As user40276 has mentioned, using Taylor’s Theorem, we can write $$ \forall \mathbf{x} \in \mathbb{R}^{n}: \quad f(\mathbf{x}) = f(\mathbf{a}) + \sum_{i = 1}^{n} \frac{\partial f}{\partial x_{i}} \Bigg|_{\mathbf{x} = \mathbf{a}} \cdot (x_{i} - a_{i}) + R(\mathbf{x}), $$ where $ R: ...


2

In $\mathbb R^n$, the norm $\|\cdot\|_\infty$ is the one whose unit ball is the $n$-dimensional cube $[-1,1]^n$. Thus $\|x\|_\infty$ is how much you have to scale that cube so that $x$ will lie on its surface. For $\mathbb C^n$, the definition is formally the same, though we might not be happy calling the unit ball a "cube". For $m\times n$ matrices, I ...


2

If you add $f(0)=0$ then you get $f(\lambda x)=\lambda f(x)$ for $\lambda\in[0,1]$ and then $$0=f\left(\frac{1}{2}x + \frac{1}{2}(-x)\right) = \frac{1}{2}f(x)+\frac{1}{2}f(-x)$$ so $$f(-x)=-f(x).$$ Next, show that $f(ax)=af(x)$ for all $a>1$. This shows it is actually true for all $a\in\mathbb R$. Then you can show that ...


2

In general, no. Take $U = (0,1) \subset \mathbb{R}^1$, $p=2$ and $F(z,w,x) = z^4$. Then take $$w_n(x) = \begin{cases} n^{1/4} x, & 0 < x < n^{-1} \\ n^{-3/4}, & x \ge n^{-1} \end{cases}$$ so that $w_n' = n^{1/4} 1_{(0, n^{-1})}$. It's easy then to see that $w_n' \to 0$ and $w_n \to 0$ in $L^2$, so $w_n \to 0$ in $W^{1,2}(U)$. Taking $w = ...


2

If you want to avoid the subsequence argument you can use Chebyshev's inequality. For any $t > 0$ and $n \in \mathbb N$ you have $$\mu(\{|f_n - f| > t/2 \}) \le \frac{2^p}{t^p} \|f_n - f\|_p$$ and $$\mu(\{|f_n - g| > t/2 \}) \le \frac{2^{p'}}{t^{p'}} \|f_n - g\|_{p'}$$ so that $$\mu(\{|f - g| > t\}) \le \frac{2^p}{t^p} \|f_n - f\|_p + ...


2

This is not true: Consider the function $f(x)=\chi_{\mathbb{R}\setminus(-1,1)}(x)|x|^{-1/2}$. Then for $p\leq 2$ we have $\| f\|_{p}=\infty$ and for $p>2$ we get $$ \| f\|_p = \left(\frac{4}{p-2}\right)^{1/p}. $$


2

If $rank(A) < n$, the strictness of $S$ may be lost, since it is now possible that $x \ne y$ but $S(x) \cap S(y) \ne \emptyset$. That is because there is $w \ne 0$ such that $Aw = 0$. You should be able to construct a counterexample to strict monotonicity using that information.



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