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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 ...


Let $Af = \int_{0}^{x}f(t)dt$ in $L^{2}[0,1]$. Then $A : L^{2}\rightarrow L^{2}$ is bounded. Let $W$ consist of all continuously differentiable $g \in L^{2}[0,1]$ for which $g(0)=g(1)=0$. $W$ is dense in $L^{2}[0,1]$ because $\{ \sin(n\pi x) \}_{n=1}^{\infty}\subset W$ is an orthogonal basis of $L^{2}[0,1]$. However, $A^{-1}W$ is not dense because $f \in ...


For any $(a,a')\in Y$, we have that $(0,-a')\in Z$. So $(a,0)\in Z$. In other words, $Y+Z$ contains the subspace $W=\{(a,0):\ a\in C^1[0,1]\}$. So now we need a Cauchy sequence in $W$ that is not convergent in $W$. For instance $\{(a_n,0)\}$, where $a_n(t)=(t+1/n)^{1/2}$.


The symbols $\langle x^*,v_n\rangle$ just express $x^*(v_n)$, the functional $x^*$ evaluated at $v_n$. It is a common notation, inspired in the Hilbert space case, where the dual is the same original space.


Maybe it will be useful to consider an example of two norms $F$ and $G$ of a vector space $X$ not being equivalent to each other. What it means is that at least one of the quantities $\sup\limits_{x \in X}\frac{F(x)}{G(x)}$ or $\sup\limits_{x \in X}\frac{G(x)}{F(x)}$ is unbounded, i.e. there is a sequence $(x_n)_{n \geq 0}$ of vectors in the space such that ...


Suppose you have a sequence which converges in $G $. The lower bound implies it converges in $F $ to the same limit. Suppose you have a sequence which ddoes not converge in $G $. The upper bound implies it does not converge in $F $. That's all you need, since metric spaces are sequential spaces.


You can start with "the norms induce the same topology". Then use the fact that a linear transformation is continuous if and only if it is bounded. And this is one of your inequalities. For the other direction, use the inverse of that linear transformation.

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