Reverse triangle inequalityHilbert spacesBanach spacesBochner integralRecent reverses for the discrete generalised triangle inequality and its continuous version for vector-valued integrals in Banach spaces are surveyed. New results are also obtained. Particular instances of interest in Hilbert spaces and for...
Using the reverse triangle inequality, we conclude \({{\mathbb {E}}[\,|Q-{{\mathbb {E}}[\,Y\,]}|\,]}=0\), thus \(Q={{\mathbb {E}}[\,Y\,]}\) is constant. If \({{\mathbb {E}}[\,Y\,]}\le \nicefrac {1}{2}\), then \({{\mathbb {P}}[\,Q\le \nicefrac...
\end{aligned} But since \Gamma _t is analytic, it follows from (3.1) and (3.2) and the reverse triangle inequality that \begin{aligned} \left| \int _{\Gamma \cap D}\textrm{d}z\right| \le (1+\varepsilon )|\partial D|. \end{aligned}...
Suppose that \{ a_n \} is a sequence such that \lim_{n \rightarrow \infty} a_n = L . Then, use the reverse triangle inequality | |x| - |y| | \leq | x - y | to an \epsilon/N proof that \l How to show a set is not finite?
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Thefirsttoconsidertheproblemofobtainingreversesforthetriangleinequality inthemoregeneralcaseofHilbertandBanachspaceswereJ.B.DiazandF.T. Metcalf[1]whoshowedthatinaninnerproductspaceHovertherealorcomplex numberfield,thefollowingreverseofthetriangleinequalityholds (1.3)r n k=1 x k ≤ n k=1 x k , ...
, at which point Theorem4follows from Theorem5after chasing definitions. (It is also possible to establish the reverse implication, but we will not need to do so here.) A remarkable fact about this theorem is that the point need not be in the support of ...
Told in reverse chronological order, Lee Chang-dong's tragic and thought-provoking drama traces the life of a broken man haunted by his past as he revisits the key moments that led to his undoing. Through its unique narrative structure, it explores themes of love, loss, and the las...
and using the reverse triangle inequality of the norm, we obtain $$\begin{aligned} \int _0^t \left\vert \ln u_\varepsilon (t) \right\vert ^2_{l^2(\varomega )} \textrm{d}t \le \int _0^t c^2 \left( \left\vert \nabla \ln u_\varepsilon (t) \right\vert _{l^2(\var...
. the reverse inequality follows from the fact that each \(t\in [0,t]\) is in \(\mathcal {t}_0\) and so \(x_t\le \underset{\tau \in \mathcal {t}_0}{\mathop {\mathrm {ess\,sup}}}\, x_{\tau }\,\,\mathbb {p}\text {-}\hbox {a.s.}\) . \(\square \) ...