# Tag Info

## New answers tagged normed-spaces

0

No. If the norm were induced by an inner product it would satisfy the parallelogram law $$||f+g||^2+||f-g||^2=2(||f||^2+||g||^2).$$Simple examples show this is not so (for example, the characteristic functions of two disjoint sets).

0

Hint: if a norm is derived from an inner product in that way, then the inner product is uniquely determined by the norm and there is an explicit algebraic expression for (f,g) in terms of ||f||, ||g||, ||f+g|| and ||f-g||.

1

Hint: take a look at the parallelogram law.

3

By definition $\|A\|=\sup_{\|x\|\leq 1} \|Ax\|$, and by Hahn-Banach theorem we can show $\forall y\in Y, \|y\|=\sup_{\|f\|\leq 1,f\in Y^*} \|f(y)\|$. Combining them, we have $\|A\|=\sup_{\|x\|\leq 1} \sup_{\|f\|\leq 1,f\in Y^*} \|f(Ax)\|$, which is what you want.

0

Hint: You don't need to find the best possible $K$; any positive $K$ that works will do. Try using the Triangle Inequality.

1

By the definition of norm, for every $\epsilon>0$ there is a unit vector $u$ such that $f(u)>1-\epsilon$. Given $x$ with $f(x)\ne 0$, let $$y = x-\frac{f(x)}{f(u)} u$$ By construction, $f(y) = 0$, meaning $y\in \ker y$. Hence, $$\operatorname{dist}(x,\ker f )\le \|x-y\| = \frac{|f(x)|}{|f(u)|}<\frac{|f(x)|}{1-\epsilon}$$ Since $\epsilon$ was ...

0

Let $X$ be a Banach space and let $\{x_d\colon d\in D\}$ be a dense subset of the unit ball of $X$. Consider the space $\ell_1(D)$ of all absolutely summable sequences on $D$. We define a linear map $T\colon \ell_1(D) \to X$ by $$T\Big((\lambda_d)_{d\in D}\Big) = \sum_{d\in D}\lambda_d x_d\qquad ((\lambda_d)_{d\in D} \in \ell_1(D)).$$ This is a well-defined ...

0

Hint: If $X,Y$ are normed linear spaces and $T:X\to Y$ is a bounded linear map, then $T$ is not just uniformly continuous, it is Lipschitz, with Lipschitz constant equal to $\|T\|,$ the operator norm of $T.$ So we only need to show $f: l^2 \to l^1$ is a bounded linear operator. Wait! We don't even know yet that $\{a_n\}\in l^2 \implies f(\{a_n\})\in l^1.$ ...

1

Let $\delta=\inf\{\|x-x_0\| :x \in K\}$, this implies that there exists a sequence $(x_n)$ in $K$ such that $||x_0-x_n||\to \delta$. Since $K$ is a closed and bounded subset of $\Bbb R^n$, it is compact so there is a convergent subsequence $(x_{n_k})$ of $(x_n)$ such that $x_{n_k}\to x^*\in K$. We then have $$||x^*-x_0||=\lim_{k\to \infty}||x_{n_k}-x_0||= ... 0 a) Let w be an arbitrary point of A+B. Then w has the form a+b with a in A and b in B. A+B is open if each of its points has a neighbourhood contained entirely in A+B. We know that there should be neighbourhoods N_a and N_b of a and b contained entirely in A and in B, respectively, because these two sets are open. Can these ... 2 Define the norm \lvert \cdot \rvert_1 \, \colon X \rightarrow [0, \infty) by \begin{equation*} \lvert x \rvert_1 =\sum_{l=1}^{n} \lvert \lambda_l \rvert, \end{equation*} where x \in X is given by x = \sum_{l=1}^{n} \lambda_l v_l. Let us now choose a vector x_i from the sequence \{x_i \} . Then, by definition of \lvert \cdot \rvert_1, ... 2 Not in general. Here is a counterexample. Let X = C([-1,1]). For t \in [-1,1], let \delta_t denote the point mass / evaluation functional \delta_t(x) = x(t). Let D = \{ \delta_q : q \in [-1,1] \cap \mathbb{Q}\}. Then D is countable and we have \|x\| = \sup_{f \in D} |f(x)| for every x \in X. Let$$y(t) = \begin{cases} 4t, & -1 \le t ...

1

Is there a typo on your right hand side? I assure you mean $f(||y||-||x||)$. If so, every $f(\cdot)$ that is monotonically decreasing should do the trick. What is left to check are the limits on $x$ and $y$

6

$$||x||+||y||-||x+y|| \\ = ||x||-\frac{||x+y||}{2}+||y||-\frac{||x+y||}{2} \\ \le ||x -\frac{x+y}{2}|| + ||y -\frac{x+y}{2}|| \\ \le \frac{||x-y||}{2} + \frac{||y-x||}{2} \\ \le ||y-x||$$

0

Yes, you are right about Baire. Sketch: Let $E_n = \{f\in C^\alpha : \|f\|_\alpha \le n\}.$ Show each $E_n$ is closed in $C.$ We'll be done if we show the interior of each $E_n$ is empty. So let $f\in E_n.$ It suffices to show $f$ is the limit in $C$ of functions not in $C^\alpha.$ Consider $f +x^{\alpha/2}/m, m=1,2,\dots.$

0

No, $S$ is intersection of two closed set thus it is close, thus if $S=\mathbb{Q}$ then $\mathbb{R}=\bar{S}=S=\mathbb{Q}$ contradiction.

2

It is not necessarily true. Consider $x_n = 1/\sqrt n \not\in \ell_2$. Then for any $y_n \in \ell_2$ $$\| x_ny_n \|_2 = \sqrt{\sum_{i=1}^\infty |x_ny_n|^2} \leq \sqrt{\sum_{i=1}^\infty |y_n|^2} = \| y_n \|_2$$ That is, $x_ny_n \in \ell_2$ for all $y_n \in \ell_2$.

1

No, not in general. For instance, suppose $A=\mathbb{C}$ with the standard norm. Then (via a change of basis $(a,z)\mapsto (a+z,z)$) we can identify $A_+$ with $\mathbb{C}^2$ with coordinatewise multiplication, and your norm with $\|(a,b)\|=(|a-b|^p+|b|^p)^{1/p}$. Now consider the elements $x=(1,0)$ and $y=(1,1/2)$. We have $\|x\|=1$ and ...

1

In addition to the norms suggested in comments, here is one more (inspired by B.S.Thomson): for $t\in [0,1]$, let $$\|f\|_t = |f(t)| + \int_0^1 |f(x)|\,dx$$ The integral term is only needed to make this a norm rather than a seminorm. The fact that these are mutually nonequivalent follows by considering $f_{a,n}(x)=\max(0,1-n|x-a|)$ which satisfies $$... 1 On one side, let x \in \bar E, choose a sequence (x_n) \in E^{\mathbf N} such that x_n \to x. If now x' \in X' is given such that \Re x'|_E \ge 1. Then, as \Re x' is continuous, we have$$ 1 \le \Re x'(x_n) \to \Re x'(x) $$That is \Re x'(x) \ge 1. Along the same line we see that \Re x'|_E \le 1 implies \Re x'(x) \le 1. For the other ... 3 Hint: try checking that P([0,1]) \ni p \mapsto p' \in P([0,1]) is not bounded in the unit sphere. Here P([0, 1]) is the space of polynomials in [0,1], with the sup norm. 1 Let A:U\rightarrow V be bounded. If X \subset U is bounded there exists a K \in \mathbb{R} such that ||x|| < K for all x \in X. Let y \in A(X) i.e. Ax = y for an x \in X then ||y|| = ||Ax|| \leq ||A||\cdot ||x|| \leq ||A||\cdot K therefore A(X) is bounded too. Now let A send bounded sets to bounded sets and let K be a bound for ... 0 You can define a linear functional \theta on the one-dimensional subspace W = \mathrm{Span} \{v_0\} by \theta(\alpha v_0) = \alpha \|v_0\|. It is trivial to check that this is linear and bounded, and moreover if \|\alpha v_0\| = 1 then |\theta (\alpha v_0)| = 1 so that \|\theta\| = 1. Now extend \theta to all of V using the Hahn-Banach ... 0 It is just a weighted L^p-space. The norm is given by$$ \|u\|_{L^2_\rho} ^p:=\int_D |u(x)|^2 \rho(x) dx, $$scalar product is$$ (u,v)_\rho = \int_D \rho(x) u(x)v(x)dx. $$1 Look at your second-to-last displayed equation. We have$$ \Re f(x_3) < t \le \Re f(x_2), \qquad x_2 \in E_2, x_3 \in E_1 + U(0,r). $$That is,$$ \Re f(x_1) + \Re f(u) < t \le \Re f(x_2) \qquad x_2 \in E_2, x_1 \in E_1, u \in U(0,r) $$Now let m := \sup_{u \in U(0,r)} \Re f(u). As f \ne 0, we have m > 0. Then$$ \Re f(x_1) + m \le t \le ...

0

Your solution to the only if direction gives a good hint: Assuming we have $r, R > 0$ such that $||x|| < r \implies ||x||' < 1 \implies ||x|| < R$, let's try setting $a=1/R$ and $b=1/r$. Notice that our assumption tells us that if $||x||' < 1$, then $||x||/R = a ||x|| < 1$. My hint would be that it follows from this that $a||x|| \leq ... 3 Hint: The sequence of numbers$\| x_n\|$is real and bounded, so it must contain a convergent subsequence by the Bolzano–Weierstrass theorem. 1 Indeed!$||x_n-x_m|| < \epsilon$for$n,m > N$is equivalent to saying that$x_m \in B_\epsilon(x_n)$for$n,m > N$. 1 Hint. For$x \in V$, what is a collinear vector$y$to$x$such that$\Vert y \Vert <a$? What happens then to$\Vert y \Vert^\prime$? And vice versa? -1 According to your definition of$C$,$A\cap C =\emptyset$( if$f\in A$,$|f|$eventually becomes smaller than$\epsilon$.) Perhaps it makes more sense to consider the set:$ C=\left\{ f\in A\,\,\, |\,\,\, \exists M>0 \,\,\,s.t.\,\,\, f(x)=0\,\,\, \forall |x| \geq M \right\}. $0 I think this is true for$n=2$, even for the set$\{x \in Bd \space B(\theta,1): |f^{-1}(\{x\})|\ge2\}$. (It should follow from the fact that the boundary correspondence induced by a self-homeomorphism of open disk is monotone.) But since you asked about all dimensions, here is a counterexample for$n\ge 3$. Replace the closed unit ball by the product of a ... 1 (1) is the definition of "complete". (2) is a consequence of the definitions, but is not how "closed" is defined. Completeness is only definable for metric spaces (or certain other spaces with a similar structure). Normed and inner product spaces are just examples of metric spaces where the metric is defined in terms of other structure on the set. Closed, ... 3 A linear functional on a$\mathbb{K}$-vector space$f:E\rightarrow\mathbb{K}$is continuous if and only if it is Lipschitz, and thus if and only if there exists$M\geq 0$such that$\|f(x)\|\leq M\|x\|$for all$x$. The norm on the real Hilbert space$\ell^2(\mathbb{N}^*)$is associated to a dot product which verifies some inequality. 2 Hint:$g$is linear. A linear operator is continuous if and only if it is continuous at$x=0$, if and only if it is bounded. Try to find a constant$C>0$such that for all$x$: $$|g(x)|\leq C \|x\|$$ 6 The inner product is already determined by the norm using the Polarization identity so no need to try and build it in a non-constructive way. If you define a function$\left< \cdot, \cdot \right>$on$X \times X$by $$\left< u, v \right> := \frac{||u + v||^2 + ||u - v||^2}{4}$$ then clearly$\left<u, u\right> = ||u||^2$for all$u \in ...

1

In any normed vector space, it's impossible for a sequence to converge to two different limits. So if we believe that $\| \cdot \|_X$ really is a norm on the vector space $P([0,1])$, then "uniqueness of limits" follows. Of course, some Cauchy sequences won't converge to any limit at all, but that's a separate issue. To be more concrete, suppose that ...

2

Let's take a Cauchy-sequence $(a_{n})_{n}\subset V$. By the completeness of $V$ with respect to the first norm $\Vert\cdot\Vert_{1}$, we have $$\forall\,\epsilon>0,\,\exists\,N\in\mathbb{N}:\forall\,n\geq N:\Vert a_{n}-a\Vert_{1}<\epsilon$$ for some $a\in V$. As $\Vert\cdot\Vert_{1}$ and $\Vert\cdot\Vert_{2}$ are equivalent, there exists $C>0$ such ...

0

A space must be complete in order to be compact. Complete means that every Cauchy sequence converges, that is every sequence "approaches" to a point that belongs to the considered set. For example, the sequence $$a_n=\bigg(1+\frac{1}{n} \bigg)^n$$ is a sequence in $\mathbb{Q}$ but it does not converge into $\mathbb{Q}$. Therefore, $\mathbb{Q}$ is not ...

3

Another reason: a compact subset of the reals is closed and bounded (Heine-Borel theorem), and the rationals are distinctly not bounded.

2

A compact space is complete. Another reason: a compact subspace is closed. And precisely, the closure of $\mathbf Q$ is $\mathbf R$.

0

Take the constant function $f(x)=1$ to show that $||T||=1$.

1

For finite $p$, suppose let $D$ be a countable dense set (containing $0$) of $K$ (which, as Daniel Fischer remarked in his comment) must exist, as $l_p(\{p\})$ for a singleton set $I$ is isomorphic to $K$. Then all sequences from $D$ that eventually $0$ can be shown to be dense in $l_p(I,K)$ If $I$ is not countable, you can easily find $|I|$ many vectors ...

0

hint To show $l_p(I,K)$ is separable. First you need to find a good candidate for a countable set $S$ with closed span $l_p(I,K)$.

2

The answer to the first question is "yes", and I guess this is originally due to Whitney. For a proof see Infinitely differentiable function with given zero set? The answer to the second question is "no" in general for $n \ge 2$. E.g., if $n=2$ and $A$ is the unit circle in the plane, then you can find a regular value $y=f(x)$ where $x$ is contained in the ...

1

I'm not sure that this is generally true, but it's true for normed vector spaces over $\mathbb R$ at least. To see this you have only have to normalize the vector. Let $\alpha_1=|x|_1$ then $|x/\alpha|_1 = 1$ so $x/\alpha$ is in $B_1$ and therefore in $B_2$ so $|x/\alpha_1|_2\le1$. Which means that $|x|_2\le |x|_1$ and the other way around works too.

1

Suppose that $\|x\|_1 \le 1 \implies \|x\|_2 \le 1.$ For any $x \not= 0$ you can let $y = x /\|x\|_1$ so that $\|y\|_1 = 1$ and thus $\|y\|_2 \le 1$. In other words, $\|x\|_2 \le \|x\|_1$. Likewise, if $\|x\|_2 \le 1 \implies \|x\|_1 \le 1$, then $\|x\|_1 \le \|x\|_2$ for all $x$.

1

Often (not always) it's simpler to show the continuity of a linear functional on a normed space by explicitly giving a bound on its norm rather than showing that its kernel is closed. Here, we have $$\lvert L(\varphi)\rvert = \biggl\lvert \int_{-1}^0 \varphi(t)\,dt - \int_0^1 \varphi(t)\,dt\biggr\rvert \leqslant \int_{-1}^1 \lvert \varphi(t)\rvert\,dt ... 0 Let us answer the first question, hopefully the other one is similar enough: If T^{\ast} is surjective, then it is an open map, and so \exists r>0 such that$$ B(0,r) \subset T^{\ast}(B(0,1)) $$where these denote the open balls in their respective spaces. Hence if \varphi \in X^{\ast}, \|\varphi\| = 1, \exists \psi \in B(0,1) such that$$ ...

3

The claim that $T$ is surjective implies range of $T^*$ being closed is not true. Take $X=c_{00}=Y$ the space of sequences with finite length. The dual space can be identified with $l^1$. Define $$Tx = (x_1, x_2/2, \dots, x_n/n,\dots).$$ Clearly, $T:X\to Y$ is injective and surjective, however $T^{-1}$ is not bounded. Let $g\in l^1$ be given. Then for ...

1

Fix $y \in Y$. Let $z$ be a unique point in $Y$ such that $||y^*|| = \frac{||y^* z||}{||z||}$, ie. $$\max_w\frac{||y^*w||}{||w||} = \frac{||y^* z||}{||z||}.$$ (Note that we need completeness of $Y$ to be assured of the inclusion of $z$ in $Y$.) Since $T$ is surjective, there exists a $\tilde x \in X$ such that $T \tilde x = z$. Therefore  ||T^* y^*|| = ...

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