# Tag Info

0

Note that if $f,g\geq0$ then we certainly have $$\int_a^bf(x)g(x)\ dx\leq(b-a)\max_{x\in[a,b]}f(x)\max_{x\in[a,b]}g(x).$$ If you're unsure of why, look at the graph. In this case, $x^4,(f(x)-g(x))^2\geq0$ and so $$\int_0^2x^4(f(x)-g(x))^2\ dx\leq2\cdot2^4\|f-g\|_\sup$$ from which your inequality follows by taking square roots. Usually in this kind of proof ...

0

As usual, $\delta$ was chosen after the estimate arrived at $\le 4\sqrt2 \delta$. For the estimate, note that $(f(x) - g(x))^2 \le \|g-f\|_\text{sup}^2$, thus $$\left(\int_0^2 x^4 (f(x)-g(x))^2 \ \mathrm dx \right)^{\frac12} \le \left( \|g-f\|_\text{sup}^2 \int_0^2 x^4 \ \mathrm dx\right)^{\frac12} = \sqrt{\frac{2^5}5} \|g-f\|_\text{sup}$$ So the constant ...

0

The result is true if we assume the $E_i$ to be Banach spaces and restrict ourselves to complete norms. Let $\|\cdot\|_{\rm alt}$ be a complete norm on $E:=E_0\times\dots\times E_{n-1}$ with $\|(0,\dots,0,x_i,0,\dots,0)\|_{\rm alt}=\|x_i\|$ for $x_i\in E_i$. Then $\|(x_0,\dots,x_{n-1})\|_{\rm alt}\le\|(x_0,\dots,x_{n-1})\|$ where the latter denotes the ...

2

The statement about continuity is true: If $f$ is additive, the difference $f(x+h)-f(x)$ is independent of $x$, hence continuity can be tested at one single point. Furthermore if $f$ is defined on the real numbers, additive and continuous at a single point, it is continuous everywhere, hence determined by the values $f(x), x \in \mathbb Q$. In particular it ...

0

Seems like the result is not true for normed spaces. For any linear space $E$ with base vector set $B$ we have $\|\cdot\|_{B,\infty}:=\sum_{b\in B}\lambda_b b\mapsto\max_{b\in B}|\lambda_b|$ is a valid norm on $E$. If we look at the space $c_{00}\times c_{00}$ with the standard base $B:=\{(e_i,0)\}_i\cup\{(0,e_i)\}_i$ as well as the non-standard base ...

1

No. For instance the set $\bigl\{{n\over n+1}e_n\mid n\in\Bbb N\Bigr\}\cup\{0\}$, where $e_n$ is the standard $n$'th unit vector is weakly compact in $\ell_2$, but has no element of maximal norm.

0

let $x_n+M$ be a cauchy sequence in $X/M$ then $x_n$ is a Cauchy sequence in $X$ and as $X$ is a Banach Space and hence complete.Thus $x_n$ converges to some $x\in X$.Hence $x_n+M$ converges $x+M \in X/M$

3

Try something like $f_n(x) = ne^{-nx}$. Then $$\|f_n\|_1 = \int_0^1 ne^{-nx} \, dx = 1 - e^{-n}$$ for all $n$ but $\|f_n\|_\infty = n$.

0

By definition, $d(u,F)$ is the infimum of distances from $u$ to the points of $F$. Since $0\in F$, we have $d(u,F)\le \|u-0\|=1$. This explains why $d$ is bounded above on the unit sphere. If $F=E$ then of course $d(u,F)$ is always $0$. Otherwise, $\sup_{\|u\|=1} d(u,F)=1$; this fact is known as Riesz's lemma. The proof can be found in many places ...

3

No. Not every normed space has an inner product which gives rise to the given norm. A normed space $(V, \|\cdot\|)$ is an inner product space if and only if it satisfies the parallelogram law: $\|x+y\|^2 + \|x - y\|^2 = 2(\|x\|^2 + \|y\|^2)$ for all $x, y \in V$. An example of a normed space which does not satisfy the parallelogram law, and is therefore ...

1

If Y is a banach space, than the claim is true, using baire category on Y and hann banach theorem, to show that every weakly convergent sequence is bounded. Proposition: if for any functional $\varphi \in Y^*$, $\varphi \circ L$ is bounded, than L is bounded. proof: let $(x_n) \subset X, \|x_n \| \to 0$, than for any $\varphi \in Y^*$, $\varphi \circ ... 1 People define isometries differently - some specify that an isometry$f$is always bijective. I will not assume this, but if you do, you need only add the condition "$f$is surjective". One sufficient condition would be that$f$is a linear map. If this is the case, then for any$x,y\in X$, we have ... 0 I think Alex actually intends to prove ||f|| is certainly a norm. Just one comment: Use$||f+g||^2 $to avoid square roots.$||f+g||^2 = ||f||^2 + ||g||^2 +2 |<f,g>| \leq ||f||^2 + ||g||^2 +2 ||f||||g|| = (||f||+||g||)^2, and we are done. 0 There's no reason to do all these calculations. The norm induced from any inner product obeys the triangle inequality as a consequence of the Cauchy-Schwarz inequality, so just state that your norm is induced from the inner product \begin{align} \left<f, g\right> := \int_{0}^{1} f(x)\overline{g(x)} \, \mathrm{d}x. \end{align} 0 We do have the following result, valid for finite dimensional normed vector spaces over normed and complete fields. Considerk$is a field with a norm$|\cdot | \colon F \to [0, \infty)$that makes him a complete normed field ( the norm is sub-additive and multiplicative). Moreover,$V$is a finite dimensional vector space over$k$that has a norm ... 0 It seems the following. Suppose that there exist points$x$and$y$in a normed space$X$such that$\|x\|\ge 1$,$y\in C$and $$\|x-y\|<\|x-P_C(x)\|.$$ Then Triangle inequality implies $$\|x\|\le \|x-y\|+\|y\|<\|x-P_C(x)\|+1=\|x-P_C(x)\|+\|P_C(x)\| =\|x\|,$$ a contradiction. 1 Consider the sequence$(f_n)_{n \geq 0}$with $$f_n(x) = \sum_{k=0}^n 2^{-k} \chi_{[1-2^{-k+1},\, 1-2^{-k})}(x)$$ where$\chi$denotes the indicator function. If I got the indices right these should be step functions where$f_{n+1}$differs to$f_n$by a new step of length$2^{-(n+1)}$and height$2^{-(n+1)}$. Every$f_n$is piecewise continuous and they ... 0 It seems the following. If$r=0$then$\partial D(a,r)=\{a\}$. Suppose that$r>0$and$b\in D(a,r)$. Let$c\in D(b,r/2)$. Since$(\mathbb{Q},|\cdot|_p)$is an ultrametric space, we have$|a-c|_p\le\max\{|a-b|_p,|b-c|_p\}=r$. That is$c\in D(a,r)$. So$D(a,r)$is a clopen set, that is$\partial D(a,r)=\varnothing$. -2 It is true. It follows the condition of optimality for optimization problem. Proof: Consider$F(y) = \left<x-y,x-y\right>$. The problem $$F(y)\to\min,\ y\in M$$ has a unique solution,$P(x)$. Since$F$is a convex function and$M$is a convex set, then the following holds: $$\left<\nabla F(P(x)), v - P(x)\right> \geqslant 0\text{ for all ... 0 Given n,m\in \mathbb N; \exists p\in \mathbb N such that ||a_n-a_m||<\epsilon \forall m,n\geq p Now ||T(a_n)-T(a_m)||=||T(a_n-a_m)||<\epsilon 2 Surely, a projection being a linear continuous map, is Lipschitz, and so uniformly continuous. Now uniformly continuous maps take Cauchy sequences to Cauchy sequences. 1 No, this isn't even true in general for norms induced by inner products: Consider \mathbb{R}^2, and the decomposition \mathbb{R}^2 = X \oplus Y into the x- and y-axes, so that the projections of (x, y) onto X and Y are respectively (x, 0) and (0, y). Now, consider the inner product given in the standard basis by$$\langle (x, y), (x', y') ... 3 No. Consider the real plane, and the subspaces$y = 0$and$y = x/10$. The projection of$v = (0, 1)$along the first subspace (onto the second) is$(10, 1)$. On the other hand, if the two subspaces are orthogonal, then the projections along each space, onto the other, are indeed shorted than the original vector. 1 You want sequences very close to zero in the$l_2$norm but with$l_1$norm$1$. The freedom that$c_{00}$offers is equivalent to$\mathbb{R}^n$with no restriction on$n$. Consider the element $$p_n=(\frac{1}{n}, \ldots, \frac{1}{n}, 0,0,\ldots)$$ with$n$components equal to$\frac{1}{n}$and the rest zero. We have$||p_n||_1=1$and$||p_n||_2 = ...

1

If $\|\cdot\|$ is a norm and $A$ is invertible, then $n(x) = \|Ax\|$ is also a norm. In your case, $Ax = ({1 \over 3} x_1, {1 \over 2 }x_2)^T$. The triangle inequality follows from the original norm as in: $n(x+y) = \|A(x+y)\| = \|Ax+Ay\| \le \|Ax\|+\|Ay\| = n(x)+n(y)$. The other conditions follow as well: $n( \lambda x) = \|A (\lambda x)\| = |\lambda ... 1$Y\cap Z$has finite codimension in$Y$and$Z$too, so you can choose$y_1,\dots,y_n\in Y$such that$Y=(Y\cap Z)\oplus \langle y_1,\dots,y_n\rangle$(where$n:=\text{dim}\frac{Y}{Y\cap Z}$) and similarly there are$z_1,\dots,z_m\in Z$such that$Z=(Y\cap Z)\oplus\langle z_1,\dots,z_m\rangle$. Now we have$m=n$(why?), so there is a linear isomorphism ... 0 This addresses a slightly different question, but I think it is the case that the only real normed spaces in which the parallelogram law$\|x+y\|^{2} + \|x-y\|^{2} = 2(\|x\|^{2} + \|y\|^{2})$holds are inner product spaces. 6 As a very beginning, this is true for all real normed spaces of dimension 2. (For dimension 0 and 1 it is trivial.) Let$\|\cdot\|$be any norm on$\mathbb{R}^2$and choose any$x$with$\|x\|=1$. Let$\gamma : [0,1] \to \mathbb{R}^2 \setminus \{0\}$be any continuous path connecting$x$to$-x$that avoids 0. Set$f(t) = \left\| x + ...

0

If the two norms satisfy $\mid x\mid_1=C_2\mid x\mid_2$, then $\mid x-a \mid_1 <d$ if and only if $\mid x-a \mid_2 <d/C_2$. Hence the open balls are related by $B_1(a,d)=B_2(a,d/C_2)$, with $B_i(a,d)=\{ x\in K:\mid x-a\mid_i<d \}$ for $i=1,2$. Thus the basis of open neighborhoods of $a$ for $\mid \cdot \mid_1$ and $\mid \cdot \mid_2$ are ...

0

Just for organization and taking the question out of the unanswered list: as pointed by Janko Bracic in the comments, the result is valid for an arbitrary metric space. The existence of the ${\bf x}_n$ is assured by the definition of infimum only. And my conclusion in the end should have been ${\rm d}({\bf a},{\bf b}) \leq r$ instead of $= r$, it was a ...

0

This link might help. Have a think about what it means to 'set the same topology'... http://math.mit.edu/~stevenj/18.335/norm-equivalence.pdf

1

If you formulate it that generally, no, this is not true. It holds for finite dimensional vector spaces over $\mathbb{R}$ or $\mathbb{C}$. But in the field itself we already have that closed and bounded implies compact. So if we work over the field $\mathbb{Q}$, then this is itself a one-dimensional vector space over itself, in the standard norm $|\cdot|$. ...

1

I think you've already basically worked out the logic. The case for $l_1$ can be easily generalised. Suppose $\{ x_n \}_{n \in \mathbb{N}} \in l_p$, then $(\sum^{\infty}_{n=1}|x_n|^p)^{1/n}<\infty$ and so $\sum^{\infty}_{n=1}|x_n|^p<\infty$. Therefore there exists an $N$ such that for arbitrary $n>N$, $|x_n|^p<1$. i.e $|x_n|<1$. And so for ...

1

A face is just one of the 'outside boundaries' of a convex set, or the whole convex set itself. You can see this from the definition as follows: If there is a single point $p\in F$that is not on the boundary, then we can take a point $k\in K$, make a line through $p$, and then all points on this line on the opposite side of $p$ will also be in $F$, by the ...

2

Yes. Let $x,y$ in the closure and $z = \alpha x + (1-\alpha)y$ with $0\leq \alpha\leq 1$. The point $x$ (resp. $y$) is a limit of a sequence $(u_n)_n$ (resp. $(v_n)_n$) of points of $A$, and $z$ is limit of the sequence $(w_n)_n$ with $w_n = \alpha u_n + (1-\alpha) w_n \in A$ by convexity of $A$. Therefore $z$ is limit of a sequence of points of $A$, and is ...

2

In the first line of the proof, $\| x^*\|$ means the regular norm. We want to see that $\| m^*\|=\| \sigma(m^*)\|$. Till the use of Theorem 3.3 we have seen that $$\| m^*\|\leq \|\sigma(m^*)\|\quad \text{and} \quad \|\sigma(m^*)\|\leq \| x^*\|, \tag1$$ where $x^*$ is any extension of $m^*$. Let $p:X \to [0,\infty)$ be defined by $p(x)=\| m^*\| \| x\|$. ...

0

Now that @kobe has demonstrated that the operator is a contraction, this justifies the search for the unique fixed point of this operator. This can be achieved by assuming that $f$ can be represented by a power series $f(x) = \sum_{n=0}^\infty a_n x^n$. It is clear that $f(0)=1$, so $a_0=1$. Now if you plug in this series into your equation, and then compare ...

1

Define $\|f\|_\infty := \max_{x\in [0,1]} |f(x)|$, for all $f\in C[0,1]$. Given $f, g\in C[0,1]$ and $x\in [0,1]$, $$|Tf(x) - Tg(x)| = \left|\left(\frac{1}{2}xf(x^2) + 1\right) -\left(\frac{1}{2}xg(x^2) + 1\right)\right| = \frac{|x|}{2}|f(x^2) - g(x^2)| \le \frac{1}{2}\|f - g\|_\infty.$$ Thus $$\|Tf - Tg\|_\infty \le \frac{1}{2}\|f - g\|.$$ So $T$ is a ...

0

Actually, there is a theorem stating that if $V$ is a finite dimensional vector space over $\mathbf{R}$, then all norms on $V$ are equivalent. This implies that in $V$ the compact sets are exacted the closed and bounded sets, as the unit sphere for instance, yes.

2

Let $\alpha = -(x,y)/(y,y)$, assuming $y \ne 0$. Then, by assumption, $$\|x\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2}$$ Using the Pythagorean Theorem: $$\left\|\left(x-\frac{(x,y)}{(y,y)}y\right)+\frac{(x,y)}{(y,y)}y\right\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2} \\ ... 1$$||x+\alpha y||^2=\langle x+\alpha y,x+\alpha y \rangle= ||x||^2+|\alpha|^2||y||^2+\bar{\alpha} \langle x, y \rangle+\alpha \langle y, x \rangle.$$Show that if \langle x, y \rangle\neq 0 there is \alpha  such that this expresion is smaller than ||x||. Clearly, if \langle x, y \rangle= 0 then$$||x+\alpha y||^2=||x||^2+|\alpha|^2||y||^2\geq ...

0

Let $L$ not be a maximal chain in the lattice of all subspaces of $X$. $M \notin$ Lat K for some M subspace of X, i.e. $Kx \notin M$ for some $x \in M$. Suppose that $\{x_n\}$ is a sequence in $M$ such that tends to $x$ in $M$ weakly. By Prop. VI.3.3 [A course in Functional Analysis, John B. Conway] $K$ is completely cont., so that $\{ K{x_n}\}$ converges ...

1

The completeness properties associated with Banach spaces and Hilbert spaces are not very relevant: a norm on a real vector space is called euclidean if it is induced by an inner product (so a Banach space is euclidean iff it is a Hilbert space). The $p$-norms on $\mathbb{R}^n$ for $p$ other than $2$ can be seen not to be euclidean in lots of ways, e.g., ...

4

It's possible to prove that norm comes from inner product if only if Parallelogram law holds, that means: $$2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2$$ For all $x,y \in \mathbb{R}^n$. For example, let $x=(1,0,0,0,\ldots)$, $y=(0,1,0,0,\ldots)$ , then: $$\|x\|^2=\|y\|^2=1$$ $$\|x+y\|^2=(2)^{\frac{2}{3}}$$ $$\|x-y\|^2=(2)^{\frac{2}{3}}$$ So Parallelogram ...

1

No, an easy counterexample for $n=2$ is $x^k = (2^{-k},0)$ for even $n$, and $x^k = (0,2^{-k})$ for odd $n$ with $\| \cdot \|_a$ being the standard norm. Then your assumption is satisfied with $\alpha = 1/2$. Now with the norm $\|(x_1, x_2)\|_b = \sqrt{x_1^2 + 4x_2^2}$ you have $\|x^k\|_b = \|x^{k+1}\|_b$ for all even $k$.

-2

Let H be a hilbert space then for $t>0$ \roh_H(t)=\sup\left{\frac{t ebsilon{2}-1+(1-ebsilon^2\frac{4})^{1/2}:ebsilon\ge 0 and less of 2}=\leq\sqrt{1+t^2}-1.and know \rho_X(t)=\sup\left{\frac{t\ebsilon\frac\2-\delta_X(ebsilon) :ebsilon between 0 and 2} and delta_X(ebsilon)={1-ebsilon^2\frac{4})^{1/2}}

1

Without working with a basis explicitly, one can argue as follows. Let $V=T(X)$. Let $R:Y^*\to V^*$ be the restriction operator, namely $R\phi = \phi_{|V}$. The range of $R$ is finite-dimensional, since $V^*$ is finite-dimensional. By definition of $T^*$, we have $T^*\phi = \phi\circ T = (R\phi)\circ T$. Thus, $T^*$ is the composition of finite-rank operator ...

0

EDIT: I just realized that you do not require $V$ to be complete (is this intentional?). In this case, my argument below breaks down ($U$ is not necessarily closed). But if $V$ is complete, then my argument shows that $W$ has to be complemented. For incomplete $V$, I am currently looking for a counterexample. It is indeed necessary that $W$ is complemented. ...

2

The point is that for every $x\in X-\{0\}$ there is $\phi\in X^\ast$ with $\phi(x)\neq 0$. I do think that one does need the Hahn-Banach theorem (more precisely a corollary, which allows one to extend continuous linear maps), as you need a continuous linear form. There is a certain analog in linear algebra, which does not need the Hahn-Banach theorem, but ...

2

It works out the same way as in the real case (and the operator norm is 1) -- we just have to be a little more careful than usual. The following holds whenever $A$ is a normed real vector space, and $A^2=A\oplus A$ is equipped with the derived 2-norm: Derive an inner product from the norm on $A^2$ in the usual way through the polarization identity: ...

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