Let $\mathscr{H}$ be a separable Hilbert space and let $T\in B(\mathscr{H})$ be such that $\displaystyle \sum_{j=1}^\infty\langle T\xi_j,\xi_j\rangle$ is absolutely convergent for any choice of orthonormal basis $\{\xi_j\}$. Does this imply that $T$ is trace-class?  The converse is true and easy to prove. It is a question from Conway's book on Operator theory.
This is a variation of the  questioned some time ago.  For a complex Hilbert space $H$ let $T: H \rightarrow H$ be a bounded operator. We call that $T$ is a trace-class operator if the following sum  $$\sum_{i}{\langle |T|e_{i}, e_{i} \rangle} &lt; \infty$$  converges, where $|T| = (T T^{*})^{\frac{1}{2}}$ is the absolute value of the operator.  Assume that  $$\sum_{i}{\langle Te_{i}, e_{i} \rangle}$$  converges for any basis in the space $H$. How to prove that if the aformentioned property holds then the operator is a trace class operator?  The progress on the problem is the following: Given an arbitrary bounded operator $T: H \rightarrow H$, one can use the following decomposition $$T = \big( \frac{T + T^{*}}{2} \big) ^{*} + i \big( \frac{T^{*} - T}{2i} \big) ^{*}$$  The latter line gives the decomposition $$T = A + i B$$ where $A, B$ are normal operators.  For the normal operators we can apply the spectral theorem that proposes that $T$ is unitary equivalent to  $$(UT U^{-1})(f(x)) = g(x) f(x)$$  where  $$U: H \rightarrow L^{2}(X, \mu)$$ $$g \in L^{\infty}(X, \mu)$$  Though this decomposition classify the operator in a broad sence, i see no direct way to conclude the statement. Are the any hints that may extend the previous argument? If not, are there any ways to conclude the statement?
Inequality involving rearrangement: $ \sum_{i=1}^n |x_i - y_{\sigma(i)}| \ge \sum_{i=1}^n |x_i - y_i|. $ Integral vanishes on all intervals implies the function is a.e. zero I am having trouble with the following problem:      $f:\mathbb{R}\to \mathbb{R}$ is a measurable function such that for all $a$:   $$\int_{[0,a]}f\,dm=0.$$   Prove that $f=0$ for $m$ almost every $x$ (here $m$ is the Lebesgue measure).    I have no problem proving this for $f$ non-negative, or under the assumption that $f$ is integrable. But the question only assumes that $f$ is measurable and no more.  My idea was the usual thing; we look at the set of points where $f$ is positive and negative and assume one of these has measure greater than zero. Then I wanted to estimate one of these by an open set, look at the integral on the open set and show that it had to be greater than zero, a contradiction. But a key part of this attack is the assumption of the absolute continuity of the integral, which only holds in the case where $f$ is integrable.  Alternatively, if it were integrable one could simply estimate $f$ by a continuous function, where the result is quite obvious.  Ultimately we are going to show that $f$ is integrable, but it is not clear to me how to show this before showing it is zero a.e. So there must be a simpler way. Does anyone have suggestions? Is there a way to set an "Adventure Mode Boundary" in Minecraft? Is there a way to set an "Adventure Mode Boundary" in Minecraft?  I am trying to create this village where players can spawn in, and then set out to explore the world and build and do normal minecraft stuff. I don't want them to destroy this village, so I want them to be in adventure mode when they are in the village and then switch them to survival when they leave, and vice-versa. When they return to the village, they get switched back to adventure mode.  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