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Yes, $T_n$ converges strongly to zero by the Riemann-Lebesgue lemma: For any $x\in H$, $$\lim_{n\to \infty} \langle x,e_n\rangle = 0$$ which itself follows from Bessel's inequality $$\sum_{n=1}^{\infty} |\langle x,e_n\rangle | ^2 \leq \|x\|^2$$ Furthermore, $T_n$ does not converge in norm, because if $n\neq m$ $$\|T_n - T_m\| \geq \|T_n(e_n) - ... 2 Let e_k = (0,0,0,\ldots,0,0,1,0,0,\ldots) where the 1 is in the kth place and all other entries are 0. This belongs to your proposed space. The set \{e_k : k=1,2,3,\ldots\} is linearly independent. 2 Part 2 is easier to answer: no, the way monotone operators are defined in functional analysis, -F is not in general monotone when F is. (This is unlike the concept of monotonicity in real analysis). Monotone operators correspond to (non-strictly) increasing functions. The reverse inequality defines dissipative operators. Part 1, geometric ... 2 As you have defined:$$ p_nf = \sum_{k=1}^{n}\left[n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(y)dy\right]\chi_{[\frac{k-1}{n},\frac{k}{n}]}(x) $$The linear operator p_n is an orthogonal projection operator onto the linear span of the orthonormal set \left\{\sqrt{n}\chi_{[\frac{k-1}{n},\frac{k}{n}]}\right\}_{k=1}^{n}. Hence, p_n^2=p_n and \|p_nf\| \le ... 1 The answer to the question in the title is no. Let H=\ell^2(\mathbb N),$$ N=\{x\in H:\ \exists m:\ x(n)=0,\ \forall n\geq m\} $$and$$ M=\{\lambda z:\ \lambda\in\mathbb C\}, $$where$$ z=\left(1,\frac12,\frac13,\frac14,\ldots\right).  Then $N$ is dense, $M$ is closed, and $N\cap M=\{0\}$.