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7

In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric" $d(x,y)=0$ if $x=y$ and $1$ otherwise, all balls $B_r(x)$ for $r \leq 1$ are the same (they are just $\{ x \}$) while all balls $B_r(x)$ for $r>1$ are also the same ...


6

Yes, $$C([0,1]) = \bigcup_{n = 1}^\infty \underbrace{\{ f \in C([0,1]) : \lVert f\rVert_\infty \leqslant n\}}_{A_n},$$ and $A_n$ is closed for each $n$ - if $\lVert g\rVert_\infty > n$, then there is a $\delta > 0$ and a non-degenerate interval $[a,b] \subset [0,1]$ such that $\lvert g(x)\rvert \geqslant n+\delta$ for all $x\in [a,b]$, and hence ...


3

The weak topology has the property that the dual is the same as the dual with respect to the original topology, if $(E,\tau)$ is a topological vector space, and $E^\ast$ its (topological) dual, then $$(E,\sigma(E,E^\ast))^\ast = E^\ast.$$ For the topology of pointwise convergence on $C(S)$, the dual is easily described: if $\lambda \colon C(S) \to ...


3

Thank you for the comments. I was able to use them to provide an answer to When does $\sum_{i=1}^{\infty} X_i$ exist for random sequences $\{X_i\}_{i=1}^{\infty}$?. I'll restate the result from there as it provides an example: Consider the measure space $L^2([0,1])$ with uniform lebesgue measure. Define the functions (Haar functions) $f_{2^i + k}$ by ...


2

It may be the case that these open balls are actually equal to each-other despite having different radii. For instance, if we equip $\mathbb{Z}$ with the usual metric, then $B_{1/2}(0) = B_{1/3}(0)$, for instance.


2

$\int_0^1\vert f_n(x)-1\vert dx=\int_0^{1/n}\vert f_n(x)-1\vert dx+\int_{1/n}^1\vert f_n(x)-1\vert dx$ Since $f_n(x)=1$ for $x\geq\frac{1}{n}$, the second term is equal to $0$. You can also remove the absolute value in the first term since for $0\leq x\leq\frac{1}{n}$, $f_n(x)=nx\leq 1$.


2

Let's prove it. Suppose $k$ is in the kernel of $A$ and $x_0$ is a specific solution, so that $Ax_0=b$. We have $$A(x_0+k)=Ax_0+Ak=Ax_0=b.$$ Conversely, suppose that $x_0$ and $y_0$ are two solutions. Then $$A(x_0-y_0)=0$$ implying that $x_0-y_0=:k$ is in the kernel. Therefore any two solutions differ by an element of the kernel.


1

To prove what you want, it suffices to prove that if $|\lambda| \neq 1$, then $(A-\lambda)$ is bijective. I was able to prove injectivity - surjectivity seems hard to just brute force, so I will post this and let someone else come up with an elegant answer :) Injectivity: Suppose $Ax = \lambda x$, and $\lambda, x\neq 0$, then $$ \frac{1}{3}x_1 = \lambda x_0 ...


1

The claim is not true for merely continuous and bounded $g$. For instance, take $$ g(v):=\min( \sqrt{\|v\|},1). $$ Fix $x$ such that $\Pi x \ne \hat\Pi x$. Then for $s>0$ such that $\|s\hat\Pi x\|\le1$ and $\|s\Pi x\|\le1$ $$ \|g(s\hat\Pi x)-g(s\Pi x) \|= \sqrt s\left|\sqrt{\|\hat\Pi x\|} - \sqrt{\Pi x}\right|, $$ hence for $s\searrow 0$ the quantity $$ ...


1

Here is an easy argument. Let $x$ be the matrix $$ x=\begin{bmatrix}v_1&v_2&\cdots&v_n\end{bmatrix}. $$ Then $$ x^*x=\begin{bmatrix} v_1^*v_1&v_1^*v_2&\cdots&v_1^*v_n\\ v_2^*v_1&v_2^*v_2&\cdots&v_2^*v_n\\ \vdots & \vdots & \ddots & \vdots \\ v_n^*v_1&v_n^*v_2&\cdots&v_n^*v_n\\ \end{bmatrix} ...


1

Not in general. If there exists $\chi\in C_\infty(\Omega)$ such that $\chi$ is injective then Stone-Weierstrass implies that $\mathcal A(\chi)=C_\infty(\Omega)$. But for "most" $\Omega$ there is no such $\chi$, and if $\chi$ is not injective it's clear that $\chi$ does not generate $C_\infty(\Omega)$: If $\chi(a)=\chi(b)$ then $f(a)=f(b)$ for all ...


1

With a clear domain definition, and $\|\frac{1}{\Delta t}\{U_{\alpha}(\Delta t)-1_\alpha\}\varphi_{\alpha}-H_{\alpha}\varphi_{\alpha}\|=\|\frac{1}{\Delta t}\int_{0}^{\Delta t}(U_{\alpha}(t)-1_\alpha)H_{\alpha}\varphi_\alpha dt\|$, can you now better establish convergence?



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