In the middle of a proof of the Inverse Function Theorem (namely, the proof of Baby Rudin), we use the fact that if $A$ is invertible and: $$ ||B-A||~||A^{-1}|| <1$$ then $B$ is invertible. The proof for this, however, relies on the fact that the vector spaces are finite dimensional (because it concludes that $B$ […]

Let $A\subset C([0,1])$ a closed linear subspace with respect to the usual supremum norm satisfying $A\subset C^1([0,1])$. Is $D\colon A\rightarrow C([0,1]), \ f\rightarrow f’$ continuous iff $A$ is finite dimensional? If $A$ is finite dimensional $D$ is continuous of course. But is the other implication true at all? Wouldn’t something like $A:=\overline{\text{span}\{\sin{\left(t+\frac{1}{n}\right)} \ | \ […]

We say that $f:[0,1]\to \mathbb{R}$ satisfies Lusin’s condition (N) provided $$m(f(B))=0 \quad\mbox{whenever}\quad B\subseteq [0,1] \mbox{ with }m(B)=0$$ where $m$ stands for the Lebesgue measure on $\mathbb{R}$. I found in here the following definition. We say that $f:[0,1]\to X$ satisfies Lusin’s condition (N) provided $$\mathcal{H}^1(f(B))=0 \quad\mbox{whenever}\quad B\subseteq [0,1] \mbox{ with }m(B)=0$$ where $X$ is a metric […]

I found the following question in my textbook: Let $E$ be a reflexive space, and let $\lbrace x_{n} \rbrace \subset E$ be a sequence such that $\lbrace f(x_{n}) \rbrace$ converges for all $ f \in E^{*} $. Show that exists a $x \in E$ such that $x_{n} \rightharpoonup x$. I can show that $x_{n}$ bounded […]

We had in our lecture on numerical analysis the following: Let $\mathrm{Lin}(X,Y)$ be the set of all linear maps $X\rightarrow Y$. Let $A\in\mathrm{Lin}(\mathbb R^l,\mathbb R^n)$ and $B\in\mathrm{Lin}(\mathbb R^n,\mathbb R^m)$ and $\|C\|_{op}:=\max_{\|x\|\leq1}\|C(x)\|$. Then our lecturer followed $\|A\circ B\|_{op}\leq\|A\|_{op}\cdot\|B\|_{op}$. So he didn’t prove it and so I’ve tried it by my own. My attempt: $$ \|A\circ B\|_{\mathrm{op}}=\max_{\|x\|<1}\|(A\circ […]

Consider a function $F \in C^{\alpha}( \mathbb{R})$ for $0 < \alpha < 1.$ Then we can take it’s distributional derivative. We can say $f = F’ \in C^{\alpha -1}( \mathbb{R} )$. My issue is going back. Say I have a $\alpha -1 -$Hölder function $f$, then how can I “integrate” it to get a primitive […]

Let $T$ be a bounded operator on $l_2$ such that there exists $\mu$ in the spectrum of $T$ which is an isolated point of the spectrum. We know that for any $x\in l_2$ the resolvent function $R(x):\rho(T)\to l_2$ given by $R(x)(\lambda)=(\lambda I-T)^{-1} x$ is analytic, and so $\mu$ is an isolated singularity of $R(x)$. Is […]

For $f$ in $C[a,b]$ define $$|| f ||_1 =\int_a^b |f|.$$ a. Show that this is a norm on $C[a,b]$. b. Show that there is no number $c \geq0$ for which $$||f||_{max} \leq c ||f||_1 \ for \ all \ f \ in \ C[a,b]$$ c. Show there is a $c\geq 0$ for which $$||f||_1 \leq […]

why the monomials are not a Schauder basis for $C[0,1]$? $p_n(x)=x^n$ such that $(p_n)$ does not form a Schauder basis for $C[0,1]$ span$\lbrace p_n : n\ge 0\rbrace$ is dense in $C[0,1]$ by Weierstrass approxmation theorem, but I cannot figure out why they are not a Schauder basis. Could you please explain

Let $H$ be a Hilbert space and $U$ a subspace. Let $U^{\bot}$ denote its orthogonal complement. I had no trouble showing $\overline{U}\subseteq U^{\bot\bot}$. But now I’m stuck for $\supseteq$. Please could someone help me finish this argument? This is what I am trying to do: Let $x \in U^{\bot \bot}$. The goal is to construct […]

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