Let $\ell^2=\ell^2(\mathbf C)$ (with $\mathbf C$ the complex numbers) be the sequence space over the complex numbers with the canonical Hilbert basis $(e(j))_{j\in\mathbf N}$ ($e(j)=\delta_{jk}$) and let $U=\mathrm{span}_{\mathbf C}\{e(j)\ |\ j\in\mathbf N\}$. The linear operator $G$ is identified by $$G:U\to\ell^2,\qquad G(e(j))=\frac{1}{\sqrt{j+1}}\sum_{k=0}^je(k).$$ Apparently, this operator is unbounded, but I'm under the impression of having proofed the opposite, which makes no sense. I'm not sure if this is the right space for those kind of questions, but I'd like to know where the flaw in my proof is.
For an arbitrary $(x_n)\in\ell^2$ the operator $G$ is given as follows: $$G((x_n))=\sum_{n=0}^\infty x_nG(e_n)=\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{x_n}{\sqrt{n+1}}e(k)\mathbf1_{\{k\leq n\}}=\sum_{k=0}^\infty e(k)\sum_{n=k}^\infty\frac{x_n}{\sqrt{n+1}}$$ So obviously the (squared) norm is: $$\lVert G((x_n))\rVert^2=\sum_{k=0}^\infty\left|\sum_{n=k}^\infty\frac{x_n}{\sqrt{n+1}}\right|^2\leq\sum_{k=0}^\infty\sum_{n=k}^\infty\frac{|x_n|^2}{n+1}=\sum_{n=0}^\infty\sum_{k=0}^n\frac{|x_n|^2}{n+1}=\sum_{n=0}^\infty|x_n|^2\frac{n+1}{n+1}=\lVert(x_n)\rVert^2$$ and hence $\lVert G\rVert\leq1$, which implies that $G$ is indeed bounded, q.e.d.
Where is my mistake? Am I maybe not allowed to swap those sums around? If so, I don't understand why - both sums are finite, because in $U$ only a finite amount of elements is $\neq0$.
This was a weird and very persistent misunderstanding of mine. The problem lies in the assumption, that $$\left|\sum_{n=k}^\infty\frac{x_n}{\sqrt{n+1}}\right|^2\leq\sum_{n=k}^\infty\frac{|x_n|^2}{n+1}$$ Whilst this inequality is true for the absolute value (triangle inequality), it isn't true for the square. In fact: $$(|a|+|b|)^2=|a|^2+|b|^2+2|a||b|>|a|^2+|b|^2\quad\forall a,b\neq0.$$ This is even more so with infinite sums, as the square of the sum might be infinite, wheras the sum of the squares isn't (see for example the harmonic series).