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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 ...

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

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

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 $$\|a+b\... 1 You neede X to consist of vectors with all their entries positive; otherwise, \langle x,y\rangle_z is not an inner product for any z that contains a zero or a negative entry. You can actually express f using a single inner product:$$ 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

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.

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\...

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