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3

You are correct up until $B(x;\epsilon) = f^{-1}(B_\mathbb{F}(f(x_0);\epsilon'))$. All you can conclude from the statement before that is $B(x;\epsilon) \subseteq f^{-1}(B_\mathbb{F}(f(x_0);\epsilon'))$, which does not imply that $B(x;\epsilon)$ is weakly open. In general, an open ball for the norm topology in an infinite-dimensional Banach space has empty ...


3

Yes, $f(B(x_{0}; \epsilon))$ is open. It does not follow that $B(x_{0}; \epsilon) = f^{-1}((B_{\mathbb{F}}(f(x_{0}); \epsilon')$. It doesn't follow even in a finite-dimensional space. To get an idea what's happening here, say $X=\Bbb R^2$ with the euclidean norm. Say $B$ is the unit ball of $X$ and define $f(x,y)=x$. Then $f(B)$ is the open interval ...


2

No, the last assumption does not follows from the first two one. To see this consider operators $T_n f = f\left(x^n \right).$


2

Of course this is quite simple, as Jonas showed. Here's a fun way to look at it. Let $K=\{x_n'\}\cup\{x'\}$. Then $K$ is a compact metric space. Regard $x_n$ and $x$ as functions from $K$ to $\Bbb C$. Uniform Boundedness shows that $||x_n||$ is bounded, and this shows that our faamily of functions from $K$ to $\Bbb C$ is equicontinuous. And $x_n\to x$ ...


1

Note that $$ |x_n'(x_n)-x'(x)|\le|x_n'(x_n)-x'(x_n)+x'(x_n-x)|\le\|x_n'-x'\|\cdot\|x_n\|+|x'(x_n-x)| $$ and the sequence $\|x_n\|$ is bounded because it converges.


1

Consider the subspace $V_\infty$ of $\mathscr l^2(\mathbb N)$ where only a finite amount of terms in a series is non-zero. This is an infinite dimensional normed vector space. Define also the subspace $V_n$ where only the first $n$ terms of a series are non-zero. $V_n \cong \mathbb R ^n$ with the standard norm. As such there is a sequence of compacta ...


1

$Y$ is the kernel of the surjective (and continuous) map $$C^1([0,1]^n) \to \mathbb R, f \mapsto f(0).$$ In particular, $Y$ is closed (as a kernel) and of codimension $1$ (since $C^1([0,1]^n)/Y \cong \mathbb R$ is one-dimensional).


1

$A(\lambda x_1,\ldots,\lambda x_m) = \lambda^{m-j}\overline{\lambda}^jA(x_1,\ldots,x_m)= \lambda^{m-k}\overline{\lambda}^k A(x_1,\ldots,x_m)$ From the above, since $\lambda \neq 0$, we have: $$\left[ \left(\frac{\lambda}{\overline{\lambda}}\right)^{k-j} - 1\right] A(x_1,\ldots,x_m) = 0$$ Thus: $\left[ \left(\frac{\lambda}{|\lambda|}\right)^{2(k-j)} - ...


1

Consider the Banach space $X = C[0,1]$ of continuous functions on $[0,1]$, with the operator $A$ of multiplication by $x$ (i.e. $Af(x) = x f(x)$). This has spectrum $[0,1]$. Let $E$ be the subspace of $X$ consisting of polynomials. This is invariant under $A$. However, $A - \lambda I$ is never surjective as an operator from $E$ to $E$, e.g. there is no ...


1

It implies that $C\|x\|\leqslant \|Tx\|$ holds true which means that $T$ is injective and has closed range. Note that it is enough to check this only on a dense subspace and ${\rm span}\{\delta_x\colon x\in X\}$ is such a subspace.


1

The spectrum of $a$ is a compact set that does not contain $0$. So there is a disk $D$ around $0$ with $D\cap\sigma(a)=\emptyset$. Thus, on $\sigma(a)$, $f:t\longmapsto 1/t$ is continuous, so $f\in C(\sigma(a))$. Then $f(a)\in C^*(a)$ via the Gelfand transform. Or, even easier, you could check that $f$ is analytic on $\sigma(a)$, and so $a^{-1}$ belongs ...


1

Let $M=\|u\|_{\infty}$. Since $G'$ is continuous, there exists a constant $K$ such that $|G'(s)|\leq K$ for all $s\in [-M,M]$. Thus $|G'\circ u|\leq K$; it follows that $|G'\circ u|^p$ is bounded by $K^p$. The desired conclusion follows from Hölder's inequality as you explained with a little correction (you have to change parenthesis by norms): ...


1

The proof follows from the following theorem from basic calculus: Let $\{a_n\}_{n\in\mathbb N}$ be a sequence of nonnegative numbers with the property that $a_{n+m}\leq a_n \cdot a_m$ for all $n,m\in\mathbb N$. Then the limit $\lim\limits_{n\to\infty}\sqrt[n]{a_n}$ exists and it is equal to $\inf\limits_{n\in\mathbb N}\sqrt[n]{a_n}.$ Theorem about ...



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