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3

1) Take $T:l^p\to l^p,(x_1,x_2,x_3,\ldots)\mapsto(0,x_1,x_2,x_3,\ldots)$ (which is your idea for (4)). Then $T$ is one to one, because if $T(x)=T(y)$, then $(0,x_1,x_2,x_3,\ldots)=(0,y_1,y_2,y_3\ldots)$ and hence $x_i=y_i$ for each $i\in\mathbb N$. But $T$ is not onto, since there is no sequence $x\in l^p$ such that $T(x)=(1,0,0,0,0,\ldots)$. 2) Take ...


3

To avoid confusing yourself with sequences of sequences, I recommend thinking of the elements of $C_0$ as functions $f:\mathbb{N} \to\mathbb{R}$ such that $\lim_{n\to\infty}f(n)=0$. Then the claim becomes: if $(f_k)_{k=1}^\infty$ is a Cauchy sequence of functions with respect to the uniform norm, there is $f\in C_0$ such that $f_k\to f$ uniformly. The ...


2

Let $A$ be ${\mathbb C}^2$ with the pointvise multiplication and the involution $(x,y)^*=(\overline{x},\overline{y})$, i.e., continuous functions on two points. But the norm let be $\|(x,y)\|=|x|+|y|$.


1

Even if you assume all the sums are inner direct sums, it is not difficult to find a counterexample. Suppose for instance $v_1$ and $v_2$ are orthogonal. Let $A_1=B_2=\mathbb{K}v_1$ and $A_2=B_1=\mathbb{K}v_2$. Then the left hand side is equal to $A_1\oplus A_2\neq0$ but the right hand side is trivial.


1

"...from our results so far we cannot conclude that the integral is positive." As Martin pointed out in a comment, this does not mean that there are examples to the contrary, only that no proof is yet at hand.


1

Let $\varepsilon > 0$, and $x(s):= s^{\varepsilon-1/2}$. Then $\left\| x\right\|_{2}^{2} =\frac{1}{2\varepsilon}$, and $\left\| T[x]\right\|_{2}^{2} = \frac{1}{2\varepsilon^3}$. So \begin{align} \left\|T[x]\right\|_{2} = \frac{1}{\varepsilon} \left\|x\right\|_{2}, \end{align} i.e., $\left\|T\right\|_{2} \ge \frac{1}{\varepsilon}$, for every $\varepsilon ...


1

After some offline help by Daniel and thanks to the last comment, I think the riddle is solved by the following: \begin{equation} \begin{split} <\nabla(\Delta+i)^{-1}\psi, (\Delta + i)^{-1}\psi> &= <(-\Delta)(\Delta+i)^{-1}\psi, (\Delta+i)^{-1}\psi> = \\ &= <(\Delta+i)^{-1}(-\Delta)(\Delta+i)^{-1}\psi, \psi>\leq ...



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