I have the following question: Let $(x_n)$ a sequence in $X$ and $x\in X$ such that for all $F\in X’$ (the dual space of the vector space X) we have that $(F(x_n))$ converge to $F(x)$ (that is: the sequence converge weakly on $X$). Let $F:X\longrightarrow\mathbb{R}$ a continuous function. Is it true that $\quad\displaystyle\liminf_{n\to\infty}|f(x_n)|\;{\color{red}\geq}\;{\color{red}|}f(x){\color{red}|}$? Recall that: […]

Let A be a non-empty bounded sub-set of $\mathbb{R}$. Let $B\subset\mathbb{R}$, given by:$B=\{\frac{a_1+2a_2}{2}|a_1,a_2\in A\}$. Express $\sup B$ in terms of $\sup A$. My attempt: Suppose $a_1,a_2\in A$ and $b\in B$. Then $a_1 \leq \sup (A)$ and $a_2\leq \sup (A)$ So $a_1 + 2a_2\leq 3sup (A)$. This gives $\frac{a_1 + 2a_2}{2}\leq \frac{3\sup (A)}{2}$. This means that […]

Let $X$ and $Y$ be two nonempty sets and let $h:X\times Y\rightarrow \mathbb{R}$ have a bounded range in $\mathbb{R}$.Let $f:X\rightarrow \mathbb{R}$ and $g:Y\rightarrow \mathbb{R}$ defined by $$f(x)=\sup\{h(x,y):y\in Y\}$$ and $$g(y)=\inf\{h(x,y):x\in X\}$$Then can we prove that $$\sup\{g(y):y\in Y\} \leq \inf\{f(x):x\in X\}?$$

Let $X$ be a compact metric space. If $f:X\rightarrow \mathbb{R}$ is lower semi-continuous, then $f$ is bounded from below and attains its infimum. I want to prove this. This is my proof: Since $X$ is compact then it follows that $f(X)$ is compact therefore it is closed and bounded (because $f(K)$ is in $\mathbb{R}$) and […]

I am trying to understand a proof I have seen of the following theorem: $$\rho(A)=\inf_{\|\cdot\|}\|A\|.$$ I understand that to do this, the idea is to show that 1) $\rho(A)\leq\|A\|$ for any norm, and then to show that, a norm exists for which, 2) $\|A\|_{\epsilon}\lt\rho(A)+\epsilon,\,\,\, \forall\, \epsilon\gt 0$, which would then indicate that, indeed, the statement […]

Suppose $\emptyset \neq A \subset \mathbb{R} $. Let $A = [\,0,2).\,\,$ Prove that $\sup A = 2$ This is my attempt: $A$ is the half open interval $[\,0,2)$ and so all the $x_i \in A$ look like $0 \leq x_i < 2$ so clearly $2$ is an upper bound. To show it is the ${\it […]

Let $M\subset \mathbb R^2$ be the Mandelbrot set. What is $\sup\{ y : (x,y) \in M \}$? Is this known? To be more descriptive: What is the supremum of all y-coordinates of all black points in the following picture: Picture File:Mandel zoom 00 mandelbrot set.jpg by Wolfgang Beyer licensed under CC-BY-SA 3.0

Let $ \mathcal{P} \subset \mathbb{R}$, $\ \mathcal{P}\neq \emptyset $ et let $b$ an upper bound of $\mathcal{P}$ Let $a \in \mathcal{P}$ and let $n\in \mathbb{N}^*$ Show that : $$\exists\ m\in\mathbb{N} \text{ such that: } \quad a+\dfrac{m}{2^n}\geq b$$ I tired For any integer $m$, we have equivalence $a+\frac{m}{2^n}\geq b\ \Leftrightarrow\ m\geq2^n(b-a)$ Now as ${\mathbb R}$ is […]

Let $A$ be an infinite set that includes Real numbers and is bounded. Let $B$ be a set of Real numbers $x$ s.t. the intersection $A\cap[x,\infty)$ is empty or includes $finite$ number of elements. prove that $\inf B$ exists. prove or disprove $\inf B=\min B$ prove or disprove that $\inf B$ exists if we don’t […]

We are given the following definition: If a sequence $(a_n)$ is bounded from below then there is a greatest lower bound for the sequence called the infimum. $m=\inf(a_n)$ if i) $(a_n) \geq m \ \forall n \in \mathbb{N}$. ii) For each $\epsilon >0 \ \exists \ n_\epsilon \ \in \mathbb{N}$ such that $a_{n_\epsilon} < m […]

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