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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 + ...


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 $p\in F$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 ...


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). 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., ...


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 ...



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