# Tag Info

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 + ... 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 ... 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. 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 ... 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$\|f_n\|_\infty = n$. 3 Ok so let$\{y_n\}$be a Cauchy sequence in$Y$. Write$y_n=f(x_n)$for some$x_n\in X$. As$f$is an isometry,$\{x_n\}$is a Cauchy sequence in$X$and thus has a limit$x$, say. Then it is immediate that$f(x)$is the limit of$\{y_n\}$. (The result more generally holds for metric spaces. I also note that some people don't require isometries to be ... 2 Pick an arbitrary Cauchy sequence$\{y_n\}\subset Y$, and let$f\colon X\to Y$be the isometry. For each$n\geq1$,$y_n=f(x_n)$for some$x_n\in X$. We have \begin{equation*} \|y_n-y_m\|=\|f(x_n)-f(x_m)\|=\|x_n-x_m\|, \end{equation*} so that$\{x_n\}$is a Cauchy sequence in$X$. Since$X$is complete,$x_n\to x$for some$x\in X$. Therefore,$y=f(x)$is the ... 2 There is nothing to prove here:$Y$is a complete metric space even if it wouldn't be a vector space. It's all in the word "isometric". This word says that$Y$is a bijective copy of$X$whereby the distance between points is preserved. This implies that the notions of "convergence" or "Cauchy sequence" in$X$and in$Y$are the same. 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: $$\left\|\left(x-\frac{(x,y)}{(y,y)}y\right)+\frac{(x,y)}{(y,y)}y\right\|^{2} \le \left\|x-\frac{(x,y)}{(y,y)}y\right\|^{2} \\ ... 2 I would recommend that you draw the function for a general n or at least think about why the value of these two integrals may be 0 . It would be interesting to know how you came around these integrals. Was it measure theory? If so, you may take a look at the Dominated convergence theorem and how to apply it to the two functions given by (or even ... 2 Yes. Let x,y in the closure and z = \alpha x + (1-\alpha)y with 0\leq \alpha\leq 1. The point x (resp. y) is a limit of a sequence (u_n)_n (resp. (v_n)_n) of points of A, and z is limit of the sequence (w_n)_n with w_n = \alpha u_n + (1-\alpha) w_n \in A by convexity of A. Therefore z is limit of a sequence of points of A, and is ... 2 Take V=L^\infty[0,1], f_1(x)=1 for all x, and f_2(x)=0 for x \in[0,1/2], and f_2(x)=1 for x \in(1/2,1]. Another example: V=C[0,1], f_1 \equiv 1 and f_2(x)=x, the norm being sup-norm. 2 The point is that for every x\in X-\{0\} there is \phi\in X^\ast with \phi(x)\neq 0. I do think that one does need the Hahn-Banach theorem (more precisely a corollary, which allows one to extend continuous linear maps), as you need a continuous linear form. There is a certain analog in linear algebra, which does not need the Hahn-Banach theorem, but ... 2 It is not too hard to show that if V is a complete \Bbb{Q} vector space, one can extend the scalar multiplication uniquely continuously to \Bbb{R}\times V\to V, so that V is also a \Bbb{R} vector space. Hence, we assume this to begin with. Also, if the vector space is complete, it is natural to assume that the underlying field is complete too. ... 2 In the first line of the proof, \| x^*\| means the regular norm. We want to see that \| m^*\|=\| \sigma(m^*)\|. Till the use of Theorem 3.3 we have seen that$$\| 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\|. ... 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: ... 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 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 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. 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 We need to show that ||f||\geq b-a. To do this take a sequence of continuous functions, which is monotone and converges to the step function \begin{align*} x(t):=\begin{cases}1,& a\leq t\leq\frac{a+b}{2} \\ -1,& \frac{a+b}{2}<t\leq b\end{cases}. \end{align*} As an example we can use piecewise linear functions such that \begin{align*} ... 1 Here's a rather pedestrian proof. If it seems confusing, just draw the picture. You're making a function that is 1 up until some point just a bit to the left of the midpoint, and -1 from the midpoint to b. This isn't continuous, so you force it continuous by connecting the two line segments. This why we needed to give ourselves a bit of room to work with, ... 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) $$where \chi denotes the indicator function. If I got the indices right these should be step functions where f_{n+1} differs to f_n by a new step of length 2^{-(n+1)} and height 2^{-(n+1)}. Every f_n is piecewise continuous and they ... 1 You want sequences very close to zero in the l_2 norm but with l_1 norm 1. The freedom that c_{00} offers is equivalent to \mathbb{R}^n with no restriction on n. Consider the element$$p_n=(\frac{1}{n}, \ldots, \frac{1}{n}, 0,0,\ldots)$$with n components equal to \frac{1}{n} and the rest zero. We have ||p_n||_1=1 and ||p_n||_2 = ... 1 A face is just one of the 'outside boundaries' of a convex set, or the whole convex set itself. You can see this from the definition as follows: If there is a single point p\in Fthat is not on the boundary, then we can take a point k\in K, make a line through p, and then all points on this line on the opposite side of p will also be in F, by the ... 1 No, this isn't even true in general for norms induced by inner products: Consider \mathbb{R}^2, and the decomposition \mathbb{R}^2 = X \oplus Y into the x- and y-axes, so that the projections of (x, y) onto X and Y are respectively (x, 0) and (0, y). Now, consider the inner product given in the standard basis by$$\langle (x, y), (x', y') ... 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). Thep$-norms on$\mathbb{R}^n$for$p$other than$2$can be seen not to be euclidean in lots of ways, e.g., ... 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 ... 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 ... 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 ...