# Tag Info

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

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 ... 2 Other than$0$? No. Try$A = -I$. 1 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$a_n(t)=(t+1/n)^{1/2}$. 1 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. 1 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 ... 1 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.

1

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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