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0

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, $T^t$.



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