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This was sorted out in comments, so I'll add a slightly different point of view. Every bounded linear operator $T:X\to V$ has the adjoint operator $T^*:V^*\to X^*$ defined by $T^*(f)=f\circ T$. The definition shows that $\|T^*\|\le \|T\|$. (In fact, equality holds by Hahn-Banach, but we do not need this here). In the special case when $T$ is an isometric ...


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Perhaps an example will help. Let $A = \{7, 13\}$. Then $E$ is the vector space of complex-valued functions on $\{7, 13\}$. Let $a = 7$; then $N_7$ is a seminorm on $E$. Letting $a=13$, it's also true that $N_{13}$ is a seminorm on $E$. For instance, letting $f(7) = i, f(13) = 0$, we have $N_7(f) = |f(7)| = |i| = 1$, whereas $N_{13}(f) = |f(13)| = 0$. Since ...


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$N_a(f)=0$ does not imply $f(x)=0$ for all $x\in A$. It only implies $f(a)=0$.


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If $V,W$ are vector spaces and $|\cdot|$ is a norm (ort seminorm) on $W$, and $T\colon V\to W$ is linear, then $v\mapsto |Tv|$ is a seminorm on $V$


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Take $X=c_{00}$---the space of all sequences which are almost everywhere $0$ and as $x_n$---the sequence having $\frac{1}{2^n}$ on $n$-th place and $0$ elsewhere.


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The FOIL rule with column vectors $u, v \in \mathbb R^n$ tells us that \begin{align} \|u - v\|^2 &= (u - v)^T (u - v) \\ &= u^T u - u^T v - v^T u - v^T v\\ &= \|u\|^2 - 2 u^T v + \|v\|^2. \end{align} Applying this to our particular problem, we find that \begin{align} \|Y - X \beta \|^2 &= \|Y\|^2 - 2 Y^T X \beta + \| X \beta \|^2. ...


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It is clear that the completeness of the whole space implies the completeness of the unit sphere. Suppose we have that the unit sphere is complete, we want to show that the whole space is also complete. $\{x_n\}$, not necessarily on the unit sphere, is a Cauchy Sequence, so is $\{\|x_n\|\}$. We have that $\|x_n\|\to a$. If $a=0$, i.e. $\|x_n\|\to 0$ which ...


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Here is a 'constructive' proof of boundedness of $T^{-1}$: Define $\|x\|_* = \|Tx\|$. Then $\|\cdot\|_*$ is a norm on $\mathbb{R}^n$ and so is equivalent to, say, $\|\cdot\|_1$. In particular, for some $c>0$, you have $\|x\|_* \ge c \|x\|_1$, or in other words, $\|Tx\| \ge c \|x\|_1$. Now let $v=Tx$, then you have ${1 \over c} |v\| \ge \|T^{-1} v\|_1$, ...


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You gave a nice proof but the contradiction isn't (at least for me) clear. To explain it we can do as this: since the sequence $(x_n)$ is bounded ($\|x_n\|=1$) in finite dimensional space (and here we used this assumption) then by the Weierstrass theorem there's a convergent sub-sequence to say $x$ and by the continuity of the norm we have $\|x\|=1$ but by ...


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The answers follow from the standard fact that $B$ has the metric $d(x,y) = \sum_{k=1}^\infty c_kp_k(x,y)/[1+p_k(x,y)]$ where $c_k$ is any sequence of scalars tending to $0$. E.g. $c_k=1/2^k$.


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Having duly noted Martin Argerami's answer, I thought it worth mentioning that for normal $A,B$ in a $C^*$-algebra, we have $$ \|(A+B)^m\|^{1/m} \leq \|A+B\| \leq \|A\|+\|B\|=\|A^m\|^{1/m}+\|B^m\|^{1/m}. $$ Observe also that for commuting normal operators, we have the following Cauchy-Schwartz type inequality $$ \|AB x\|^2=\langle A^*Ax,B^*Bx\rangle\leq ...


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No, let $V$ be separable and $e_n$ form an orthonormal basis in $V$, set $Ae_n=\frac1n e_n$. Then $\langle Au,u \rangle>0$ since every non-zero vector has at least one non-zero coefficient in its expansion, but there is no $\alpha$ since otherwise $\|Ae_n\|\geq\alpha$ for all $n$. The first condition is sometimes called "strictly positive definite" and ...


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Let $v=\frac{5}{9}$ and $w=-\frac{5}{9}$. Then $(v,w)$ is clearly not in the first set, but it is in the second since 1) $|v+1w|=0\le1$ 2) $|v+(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)w|=\frac{5}{9}|\frac{3}{2}-\frac{\sqrt{3}}{2}i|=\frac{5}{9}\sqrt{3}\le1$ 3) ...


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The part with lemma 2 isn't necessary. Any complete subspace of a metric space is closed. So the proposition follows directly from lemma 1 and you don't need to assume that $E$ is finite-dimensional. But can you prove lemma 1? Actually even if you use just “complete subspace of complete space is closed” you don't need lemma 2 since both $E$, $F$ are ...


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I claim that $\exists M \geq 0$ such that $$ |f(x)| \leq M \quad\forall x\in E \text{ such that } \|x\| \leq 1 \qquad(\ast) $$ If not, then $\exists x_n \in E$ such that $\|x_n\| \leq 1$ and $$ |f(x_n)| > n^2 $$ Hence, take $$ y_n = x_n/n $$ Then, $y_n \to 0$ and $\{f(y_n)\}$ is unbounded. Now can you prove that $f$ is continuous at $0$ from $(\ast)$?


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Not in general, for example the infinity norm is a norm on any vector space over a totally ordered field that does not come from an inner product. E.g. for a vector $\vec{v} = (v_1, \dots, v_n) \in \mathbb{R}^n$, defined $||\vec{v}||_\infty = \max\{v_i\}$. You can check that this is a norm, and does not come from an inner product.


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Note that, in your case with the $2$-norm, you have $\|Ax-b\|_2^2 = \langle Ax-b,Ax-b \rangle $, where $\langle \cdot , \cdot \rangle$ is the euclidian scalar product. But this is not always possible, for example the $p$-norm with $1 \leq p < \infty$ defined by $$\|x\|_p := \left(\sum_{k=1}^n |x_k|^p\right)^{1/p},$$ is such that there is no scalar product ...


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Yes. Because you consider the identity map $(V,\|\cdot\|_2)\longmapsto(V,\|\cdot\|_1)$. The inequality between the norms shows that it is bounded. It is not immediately obvious to me that it is a surjection in the generality of your question. But it is in the concrete case you mention, because a C$^*$-homomorphism has closed image (when the domain is a ...


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The answer is not right. (I do not elaborate, because you only asked for confirmation.)



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