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This actually works in arbitrary Banach spaces: Note first that if $B[x,r] \cap B[y,s] \subseteq S[x,r] \cap S[y,s]$, then there's nothing to do. In particular, we can assume that $B[x,r] \cap B[y,s]$ contains an interior point of one (and, therefore, both) balls. Let $B_{1}$ be the interior of $B[x,r]$ and $B_{2}$ be the interior of $B[y,s]$. Since $B_{... 0 I try a solution: a) Put $$A=\sqrt{\frac{ 4r^2\|x-y\|^2-(\|x-y\|^2+r^2-s^2)^2}{4\|x-y\|^2}}$$ $$B=\frac{\|x-y\|^2+r^2-s^2}{2\|x-y\|^2}$$ and $$z=x+B(y-x)+Au$$ with$u$orthogonal to$x-y$, such that$\|u\|=1$. My computations, if they are correct, show that we have$\|z-x\|=r^2$and$\|z-y\|^2=s^2$. So$z$belongs to both "circles". There is a ... 1 One difficulty without the triangle inequality is that the set of open balls may fail to be a base for a topology, so there may be little,if any, relation between the "metric" and some usual, useful, topologies on the space. 1 You neede$X$to consist of vectors with all their entries positive; otherwise,$\langle x,y\rangle_z$is not an inner product for any$z$that contains a zero or a negative entry. You can actually express$f$using a single inner product: $$f(x,z)=x^Tx+\sum_{j=1}^n\frac{x_j^2}{z_j} =\sum_{j=1}^n{x_j^2}+\sum_{j=1}^n\frac{x_j^2}{z_j} =\sum_{j=1}^n\left(1+\... 1 In a metric space, the triangle inequality says that if you go from point x to point y via point z, the distance is at least as big as the distance from x to y. I find it hard to argue against that. Besides the above philosophical fact, negating the the triangle inequality in a normed space is basically negating continuity of the sum: if$$\|a+b\... 8 Assume$A$is not closed. Then there is some$x_0 $in the closure of$A$, but not in$A$. This implies that the continuous function$f(x) := |x - x_0| $assumes only positive values on$A$, but the closure of the image contains$0$, which contradicts closeness of$f(A)$. Hence,$A$is closed. 2 Your sequence of equalities should have been $$\|A_\lambda f\|^2=\lambda\int_0^1f(\lambda t)^2dt= \int _0^\lambda f(t)^2=\int _0^1f(t)^2\chi_{[0,\lambda]}^2\leq \|f\|\|\chi_{[0,\lambda]}\|=\lambda \|f\|.$$ Now, I don't know what inequality you are using, but I cannot make sense of it. And you get$\|f\|$instead of$\|f\|^2$, which is a bad sign. The ... 0 For geometric intuition, let$S$be the sphere whose diameter is the segment$(0,x)$. Since the segments$(0,Px)$and$(Px,x)$are orthogonal,$Px$lies on$S$, and similarly for$Qx$. Thus$\|Px-Qx\|\leq\|x\|$, with equality iff the segment$(Px,Qx)$is also a diameter of$S$. This happens iff the four points are coplanar and form a rectangle, which implies ... 2 The quantity$\|f\| = |f(a)| + \|f\|_{TV}$is a natural norm on$BV[a,b].$If$f\in AC,$then this norm equals$|f(a)| + \int_a^b|f'|.$Suppose$f_n$is a sequence in$AC$that is Cauchy in this norm. Then$f_n(a)$is a Cauchy sequence in$\mathbb R,$and$f_n'$is Cauchy in$L^1[a,b].$Thus$f_n(a) \to c$in$ \mathbb R$and$f_n'\to g$in$L^1[a,b].$We ... 1 If$G$have a topological complimentary in$E$(in particular if$E$is finite dimensional space) i don't think that such construction can be done, because prolonging by$0$on the complementary space, will have the same norme : $$\|\hat{T}\|=\sup_{\begin{array}{c} x\in E \\ x\neq 0\end{array}} \frac{\|\hat{T}x\|}{\|x\|}=\sup_{\begin{array}{c} x=z+y \\ y,z\... 1 Assuming A is an orthogonal linear map (i.e. A^T A = I), you are correct that \|Av\| = \|v\| does not hold in general for norms other than the 2-norm. For example, take$$A = \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ -1 & 1\end{pmatrix}$$which you can check satisfies A^T A = I, and$$v = \begin{pmatrix}1 \\ 0 \end{pmatrix}$$and compute ... 1 More generally, if \{T, S_1, S_2, \ldots\} is a uniformly continuous family of maps from metric space X to metric space Y, then \{x \in X: \lim_{n \to \infty} S_n(x) = T(x)\} is closed. This does not require completeness of either metric space. 0 Your argument is fine, besides the typo where it should be (\|S_n\|+\|T\|) instead of \|S_n+T\|. And I don't think you need for X to be complete. 3 No. For example$$||(x,y)||=|x|+|x-y|$$is a norm on$\Bbb R^2$for which this is false (consider$v_1=(2,0)$,$v_2=(3,3)$). 1 Let$A - B = F$, and suppose that$B$is diagonalizable with$B = SDS^{-1}$and$D$diagonal. Let$\|\cdot\|$be a vectorial norm such that for every$x\in\mathbb R^n$and$y$defined as$y_i = |x_i|\,\,\forall i$, then$\|x\| = \|y\|$. Under these hypotesis, the Bauer-Fike Theorem assure us that for every eigenvalue$\lambda$of$A$, there exists an ... 0 Using the dual basis, the matrix representation of$T^+$is given by the transpose,$T^t\$.