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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 $L^2[-1,3].$

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To elaborate on the suggestion in the comments: Assume $A=\langle x, y \rangle \ne 0$. After replacing $y$ with $\alpha\, y$, with non-zero $\alpha \in \mathbb C$, you can assume that $A$ is (non-zero) real. At that point, square the inequality, and expand the lhs - you may take $c$ to be real to find your desired contradiction. Hope this helps.

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I think the intended question was something like: Given $p(x)\in\mathbb{C}[x]$ with degree $n$, such that $$\int_{0}^{1}x^k p(x)\,dx=0$$ for every $k\in\{1,2,\ldots,n-1\}$, show that the roots of $p(x)$ lie in the circle $|x|\leq 1$. We may notice that $p(x)$ has $n+1$ coefficients and we have $n-1$ linear constraints on them, so the space of ...

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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 $t^{2k}(-1) = t^{2k}(1)$ ...

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Let $s_n=\sum_{k=1}^n\left<v,e_k\right>e_k$. Then $$\left<v,s_n\right>=\sum_{k=1}^n|\left<v,e_k\right>|^2$$ and $$\|s_n\|^2=\sum_{k=1}^n|\left<v,e_k\right>|^2.$$ Thus $$\|s_n\|^2=\left<v,s_n\right>$$ and hence $$\|s_n\|^2\le \|v\|\|s_n\|$$ So $\|s_n\|\le\|v\|$ and hence $\sum_{k=1}^n|\left<v,e_k\right>|^2\le\|v\|^2$. ...

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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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For a simple example to demonstrate the importance of the basis being orthonormal, consider $\mathbb{R}^2$ with the standard inner product and the basis $h_1=(1,0)$ and $h_2=(1,1)$. If $h=(0,1)$ then $$h=-h_1+h_2$$ but $\langle h,h_1\rangle=0$. By the way, a Hilbert space basis of $H$ is different from a basis of $H$ as an abstract vector space (called a ...

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\begin{eqnarray} \langle T_z^*(\alpha),x \rangle_{H}&=&\langle \alpha , T_z(x)\rangle_\mathbb{K}\\ &=& \alpha \overline{T_z(x)}\\ &=& \alpha\langle z,x\rangle_H\\ &=& \langle \alpha z , x\rangle_H\\ \end{eqnarray} So $$T_z^*(\alpha)=\alpha z$$ It's clear now that $\|T_z\|=\|T_z^*\|=\|z\|$

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