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To your first question, let $$p=\begin{bmatrix}1&0\\0&0\end{bmatrix},\ \ \ q=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix}.$$ Then $$p\vee q=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ \ p+q=\begin{bmatrix}3/2&1/2\\1/2&1/2\end{bmatrix},$$ so $$p+q-p\vee q=\begin{bmatrix}1/2&1/2\\1/2&-1/2\end{bmatrix},$$ which is ...

1

Note that the two equalities $A=AA^*A$ and $A^*=A^*AA^*$ are the same, since you can obtain one from the other by taking adjoints. Assume first that $A=AA^*A$. By multiplying by $A^*$ on the left, we get $$A^*A=(A^*A)^2.$$ It follows that the eigenvalues of $A^*A$ all satisfy the equation $\lambda=\lambda^2$, so only $0$ and $1$ are possible. Conversely,...

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No, it's not a coincidence. The definition of "$\lim (T-T_k)=0$" is "$\lim||T-T_k||=0$". No, you can't do the same for any operator. You can't show that $||T-T_k||\to0$. (Let $Tx=x$. What is $||T-T_k||$?)

2

The continuous Functional Calculus will give you want you want, under your assumptions. If $\lambda \ne 0$ is a point of the spectrum, you can construct a continuous function $F_{\epsilon}$ that is $1$ at $\lambda$ and is $0$ on the remaining part of the spectrum. Then $P_{\lambda}=F_{\epsilon}(T)$ is selfadjoint and $P_{\lambda}^2=P_{\lambda}$. Because $(z-\... 1 First, note that$X_2$is an$A$-invariant subspace, so that$A_2:X_2\to X_2$. We can show formally that the adjoint of$A_2$should be the restriction of$A^*$to$X_2$, which is again$A^*$. It then suffices to note that the image of the closed unit ball under$A_2$is a closed subset of the image of the closed unit ball under$A$and is therefore ... 1 It follows from the definition of$\sup$. If $$M=\sup\left( f(x)\ : x \in B\right),$$ then there exists a sequence$x_n\in B$such that $$M=\lim_{n\to \infty} f(x_n).$$ Apply this observation with$B=\text{unit sphere}$,$f(x)=|(Ax_n, x_n)|$and$M=\|A\|$. 1 Divide by$n_k$in the inequality $$\tag{*}\|f(x_{n_k})\| > \varepsilon + n_k \|g(f(x_{n_k}))\|, \quad k \in \mathbb{N}$$ and letting$k$going to infinity, we get, by boundedness of$\left(f\left(x_{n_k}\right)\right)_{k\geqslant 1}$, that$\lim_{k\to +\infty}g\left(f\left(x_{n_k}\right)\right)=0$. By continuity of$g$we also have that$\lim_{k\to +\...

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$\|f(x_{n_k})\|\leq \|x_{n_k}\| \|f\|=\|f\|$. This implie that from your inequality, $\|f\|\geq \|f(x_{n_k})\|>\epsilon +n_k\|g(f(x_{n_k}))\|$ since $f(x_{n_k})$ converges towards $y$, $lim_n\|g(f(x_{n_k}))\|=\|g(y)\|$. We deduce: $\|f\|\geq \epsilon+n_k\|g(y)\|$. This implies that $g(y)=0$ (since $n_k\|g(f(x_{n_k}))\|$ has to be bounded) and $y=0$ ...

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Here's the issue with your boundedness proof. Your argument boils down to the following calculation: \begin{align*} \|Bf\|_{L^2}&=(\int_0^\infty\left|\frac{1}{x}\int_0^xf(t)dt\right|^2dx)^{1/2}\\ &\leq(\int_0^\infty\left(\frac{1}{x}\int_0^\infty|f(t)|\mathbf1_{[0,x]}dt\right)^2dx)^{1/2}\\ &\leq(\int_0^\infty\frac{1}{x^2}\|f\|_{L^2}^2xdx)^{1/2}\\ &...

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