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8

Remember the definition of equivalence is that there exists numbers $c,d$ such that $c\Vert x\Vert_2 \leq \Vert x\Vert_1 \leq d\Vert x\Vert_2$. What the c and d do in the picture is to stretch or shrink the shapes you've drawn. What the inequality represents is one shape fitting inside another. What equivalence means is that I could shrink the circle for ...


6

Given the axiom of choice, every vector space has a basis (though it will be a very unnatural basis), and you are correct that infinite-dimensional vector spaces are exactly those where the basis is infinite. But this kind of basis (often called a Hamel basis) is rather useless and impossible to visualize. So, a more concrete way of thinking about it might ...


6

If a function $m$ satisfies your condition, then $m(x) = 1$ for all $x \ne 0$. Just observe the following inequality: $$2^{m(x)}||x|| = ||2x|| = ||x + x|| \le ||x|| + ||x|| = 2||x||$$


5

Thanks for the help from user zhw. I claim that $$\lim_{p\rightarrow \infty} \int_{\mathbb R^N}\Bigg\{\left(\frac{|\nabla u|}{\|\nabla u\|_p}\right)^{p-2}\frac{\nabla u}{\|\nabla u\|_p}\Bigg\} \cdot \nabla v dx = \int_K \frac{1}{m(K)}\frac{\nabla u}{\|\nabla u\|_\infty} \cdot \nabla v dx$$ where $K:= \{|\nabla u| = \|\nabla u \|_\infty\}$. We assume ...


4

If $f_n(x) = (n+1)x^n,$ then $\|f_n\|_1 = 1, \|f_n\|_2 =(n+1)/\sqrt {2n+1}, \|f_n\|_u = n+1.$


4

The relaxed condition also implies $$ \left\|\frac1\alpha \alpha x\right\|\le\left|\frac1\alpha\right|\|\alpha x\|$$ and hence $$ \|\alpha x\|\le |\alpha|\|x\|\le |\alpha|\left|\frac1\alpha\right|\|\alpha x\|=\|\alpha x\|,$$ which implies equality throughout.


4

Hint: choose any constant function $g$ on the interval $[a,b]$. Then $g \in B$ but $g$ is not the zero function.


4

(a) Consider $g(t)=1$. (b) Consider $g(t)=t^2(1-t)^2$.


3

Take $\mathbb R$ over $\mathbb R$ endowed with the absolute value norm, $v = 1$, and $v_0 = 2$. Then $$1 = |v| = |v - v_0 + v_0| \neq |v - v_0| + |v_0| = 1 + 2 = 3.$$ For inner product spaces in particular, $||x||^2 = \langle x, x \rangle.$ We have \begin{align*} ||x + y||^2 &= \langle x+y, x+y \rangle \\ &= \langle x,x \rangle + 2\langle x,y ...


3

Let me argue that the whole segment between $x$ and $y$ is in $S$. For $t\in[0,1]$ set $x_t:=x+t(y-x)$. Then $x_0=x$, $x_1=y$, $x_{1/2}$ are all in $S$ by assumption: $\|x_t\|=1$ for $t\in \{0,1/2,1\}$. By convexity of the norm, we obtain immediately that $\|x_t\|\le 1$ for all $t\in [0,1]$. Convexity not only tells us something about the function values ...


3

A necessary and sufficient condition for a subset $B$ of a vector space $V$ to the unit ball of a norm on $V$ is that $B$ is non-empty, convex and symmetric about the origin ($-B = B$). If $B = \{(x, y) : x^2-1 \le y \le 1-x^2\}\subseteq\mathbb{R}^2$, then $B$ is non-empty, convex and symmetric about the origin and so it is the unit ball of a norm on ...


3

To elaborate on my comment on Michael's answer: The symbol $\left\Vert\mathbf{u}\right\Vert$ for a vetor $\mathbf{u}$ usually stands for the norm of that vector. A norm is "a function that assigns a strictly positive length or size to each vector in a vector space" (quoted from wikipedia). Having a normed vector space enables you to talk about e.g. the ...


3

Suppose $f\in L^\infty([0,1])$ and $\|f\|_\infty> 0$. Let $E = \{x:|f(x)|=\|f\|_\infty\}.$ Then $$\lim_{p\to \infty} \left( \frac{\|f\|_\infty}{\|f\|_p}\right)^p = \frac{1}{m(E)}.$$ (If $m(E)=0,$ the conclusion is that the limit is $\infty.$) Proof: Let $M= \|f\|_\infty$. Then the expression equals $$\frac{M^p}{M^p\cdot m(E) + M^p\int_{[0,1]\setminus ...


3

What it means is not that hard: We have a sequence of real numbers $\langle x_n,y_n\rangle$ which converges to a real number $\langle x,y \rangle$. It doesn't matter what your interpretation of what an inner product "means" is. So it means that the absolute value of $\left|\langle x_n,y_n \rangle - \langle x,y \rangle \right|$ gets as small as we like by ...


2

Given the Banach spaces $\mathcal{c}_0$ and $\ell^\infty$. Consider the identity: $$T_0:\mathcal{c}_0\to\mathcal{c}_0:\quad T_0:=\mathbb{1}$$ It has No continuous extension: $$T:\ell^\infty\to\mathcal{c}_0:\quad T\restriction_{\mathcal{c}_0}=T_0$$ For the details see: Werner


2

This is true for any function f--the norm isn't allowed to take on infinite values.


2

Claim 1: The topology induced by $d_2$ is finer than the topology induced by $d$. Proof of Claim 1: It is enough to show that for every $\epsilon > 0$ we can find $\delta$ such that $$B_{d_2}(0,\delta) \subset B_{d}(0,\epsilon).$$ This turns out to be easy: let $\delta := \frac{\epsilon}{2}.$ Then if $x \in B_{d_2}(0,\delta)$, we have $$\sum_{n = ...


2

No, certainly not. Convergence in $\mathbb {R}^{\mathbb {N}}$ is equivalent to convergence is each slot. Thus $e_n$ converges to the $0$ sequence in $\mathbb {R}^{\mathbb {N}}$ (here $e_n$ is the sequence with $1$ in the $n$th slot, $0$'s elsewhere), while $e_n$ does not converge in $l^2(\mathbb N ).$


2

Your question is not very clear. If the $A^\ast$ in your question simply denotes some other matrix than $A$ (rather than the conjugate transpose), you may consider $$A_0=\pmatrix{3\\ &3},\ A=\pmatrix{4\\ &3},\ B=\pmatrix{0\\ 4}.$$ We have $\kappa(A_0)=1<\frac43=\kappa(A)$ but $\kappa([A_0,B])=\frac53>\frac54=\kappa([A,B])$. If $A^\ast$ does ...


2

In an inner product space $$ \langle v, w \rangle = \frac{1}{2}(\|v + w\|^2 -\|v\|^2 - \|w\|^2) $$ So the norm determines the inner product. The norm on a normed vector space is induced by an inner product iff the function $\langle \cdot, \cdot \rangle$ defined by the above formula satisfies the axioms for an inner product. The Jordan-von Neumann theorem ...


2

Suppose that $V$ is a vector space over $\mathbb R$, $\|\cdot\|$ is a norm on it, and $$U\equiv\{v\in V\,|\,\|v\|<1\}$$ is the open unit ball with respect to the norm. Fix $x\in V$ and $x\neq 0$. Claim 1: There exists some $t>0$ such that $tx\in U$. Proof: Since $x\neq 0$, one has $\|x\|>0$. Put $t\equiv 1/(2\|x\|)>0$. Then, ...


2

Jensen's inequality? Oh geeze, I'm a bad analyst, so I can never remember something like that. Here's a trick you might try. Suppose that there is a constant $c > 0$ so that $||f||_{L^1(\mathbb{R}^n)} \le c ||f||_{L^2(\mathbb{R}^n)}$. Let's make the change of variables $x \to \epsilon x $ for some $\epsilon > 0$. That is, let's let $y = \epsilon ...


2

The proximal operator for $\|CX\|_1$ does not admit an analytic solution. Therefore, to compute the proximal operator, you're going to have to solve a non-trivial convex optimization problem. So why do that? Why not apply a more general convex optimization approach to the overall problem. This problem is LP-representable, since $$\|CX\|_1 = \max_j \sum_i ...


2

No, and a counter-example can come from the simple cases. Take $X=\mathbb{R}^2$, $v_0=\hat{i}$ and $v=\hat{i}+\hat{j}$. In this case, $v=v-v_0+v_0=(1,1)$ and so $\|v-v_0+v_0\|=\sqrt{1+1}=\sqrt{2}$ while $\|v-v_0\|=\|\hat{j}\|=1$ and $\|v_0\|=1$. In fact I think the equal sign holds only when $v-v_0$ is a non-negative multiple of $v_0$. For the Pythagorean ...


2

The conditions can not be weakend. We will show that each one is necessary Let $\left|\cdot\right|$ be any norm on $X$ and let $x\in X\setminus\left\{ 0\right\} $. By rescaling, we can achieve $\left|x\right|=1$. For any $\alpha,\beta>0$, we have that $\left\Vert \cdot\right\Vert _{1}:=\alpha\cdot\left|\cdot\right|$ and $\left\Vert \cdot\right\Vert ...


2

Simply note that $|A|, |A^\ast|$ are positive semi-definite, self-adjoint operators which both have operator norm $\Vert A \Vert$. Hence, $\sigma(|A|) \cup \sigma(|A^\ast|) \subset [0, \Vert A \Vert]$. Since the operator norm equals the spectral-radius for self-adjoint operators, we get $\Vert A \Vert \in \sigma(|A|) \cap \sigma(|A^\ast|)$. Now, since ...


2

If $V$ is a vector space of finite dimension $n$, any collection of $n + 1$ vectors is linearly dependent. Using this property you can easily show that the vector spaces in the first posting are not finite-dimensional by proving the existence of a linearly independent set of arbitrary size. For example, let's take a look at $C([0, 1])$. For every natural ...


2

It is enough to show that when one of the norms is 1, the other is bounded between two positive constants, $m$ and $M$. It is easiest to look at the case when $||(x,y)||_\infty=1$. What does that equation tell you about $x$ and $y$? You can use this to bound $||(x,y)||_3$. In terms of the shape of $||.||_3$, you really just want to sketch the curve ...


2

I'm not sure I understand this question. The distance is defined to be the norm of the difference, and the definition is then the same as it is in metric spaces. You can get geometric intuition from Euclidean space, where $\| x - y \|$ is the length of the line segment connecting $x$ and $y$. If you replace $x$ with another sequence $y_m$, the situation is ...


2

This is a matter of convention that differs between literatures and depending on what you're doing. If you're going to use raised indices in a meaningful way, then I'd expect to see the basis written as $$ B = \{ u^i \}_{i=1}^{\infty} $$ with the index up on vectors. Then your second option for $K$ would be correct. Since you wrote the vectors in the ...



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