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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. Consider $k$ 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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