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3

The function $t\cdot \chi_{[-1,1]}(t)$ is orthogonal to each $t^{2k}$ in $L^2[-1,3],$ hence is orthogonal to the the linear span of $\{t^{2k} : k=0,1,\dots \}$ in $L^2[-1,3],$ hence is orthogonal to the closure of this linear span in $L^2[-1,3].$ Therfore this closure cannot be all of $L^2[-1,3].$

2

Let $A$ denote the linear span of $\{t^{2k}\}_{k \in \mathbb{N}}$. Then, if instead of $[-1,3]$ the domain was $[1,3]$ instead, we could use the Stone-Weierstrass theorem to conclude that $A$ is dense ins $C([1,3])$ and thus in $L^2([1,3])$ since $t^2$ separates the points of $[1,3]$ and $1 \in A$. However, for $[-1,3]$, we have $t^{2k}(-1) = t^{2k}(1)$ ...

2

Regarding convergence and completeness: For $n\in N$ let $f_n(x)=0$ for $x\leq 1/2-1/(n+2)$ and $f_n(x)=1$ for $x\geq 1/2.$ Let $f_n(x)$ be linear for $x\in [1/2-1/(n+2),1/2].$ Then $(f_n)_{n\in N}$ is a Cauchy sequence with respect to the norm $\|f-g\|=[\int_0^1|f(x)-g(x)|^2\;dx]^{1/2}.$ Let $h(x)=0$ for $x\leq 1/2$ and $h(x)=1$ for $x>1/2.$ The ...

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First, note that $Z:=Im(K)$ is a closed subspace of a Banach space and thus, itself a Banach space. Thus, $K: X\to Z$ is onto. By the open mapping theorem, $K$ is open and hence, $K$ is mapping open sets to open sets. Now, assume that $K$ is compact and take the image of the open unit ball $C:=K(B_X^\circ)$ which is open in $Z$ and relatively compact in $Y$...

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Here is an alternative proof, using as a starting point that $\ell^{\infty}$ is a Banach space. As $c$ is a subspace of $\ell^{\infty}$, it suffices to show that $c$ is closed in $\ell^{\infty}$. To see this let $\mathbf{x}=(x_1,x_2,\dotsc)\in\ell^{\infty}$, and suppose that $\{\mathbf{x}^n\}_{n=1}^{\infty}$ is a sequence in $c$ converging to $\ell^{\infty}$...

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