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1

For any two ONBs there is a unitary mapping one to the other. So, if one knows $$\|T\|_2^2 = \sum|\langle e_i, Te_i\rangle |^2$$ and wishes to have an arbitrary pair of UNBs here, one can $$\|T\|_2^2 = \|UT\|_2^2= \sum|\langle e_i, UTe_i\rangle |^2= \sum|\langle U^*e_i, Te_i\rangle |^2$$ Choose $U^*$ so that $U^*e_j=f_j$, and you have the claimed ...


1

Since $\|\cdot\|$ is unitarily invariant, $\|M\|$ depends only on the singular values of $M$ and since $M^*$ has the same singular values as $M$, $\|M\|=\|M^*\|$. It remains to use the triangle inequality.


0

Hint If every term of a sequence $(b_n)$ is in the interval $[0,1]$ and $b_k=1$ for some $k\ge 2$, then the sequence is not in $M$, but it is in $\overline M$ (this needs to be proved).


0

We denote the closure of a set $A$ by $\bar A$ and the interior by $\mathring A$. Hint: Consider the set $B_1(0)$ where $B$ denotes the open ball and $0$ denotes the sequence $(0,0,\dots)$. What is the closure of this set? What is the boundary of this set? How does this set relate to $M$?


3

Assuming $\| A^T A \|$ means the Euclidean operator norm induced by $\| x \|$ you only have inequality, not equality. For an extreme example, you can have nonzero $A,x$ such that $Ax=0$, in which case $\| A^T A x \|=0$ but $\| A^T A \| \| x \| \neq 0$.


2

The inequality does not always hold. Put $X=A^{1/2}$ and $Y=B^{1/2}$. The inequality is equivalent to $|||X^2Y^2|||\le|||XY|||^2$ for every pair of positive definite matrices $X,Y$ and for every unitarily invariant matrix norm $|||\cdot|||$. Now, take $X=\pmatrix{2&0\\ 0&1},\ Y=\pmatrix{2&1\\ 1&1}$ and the operator norm (induced 2-norm), we ...


1

1) Yes. 2) Let $B_i(x,r)$ be the open ball centered at $x$ of radius $r$ for the metric $d_i$. Then for all $x\in X$, for all $r>0$ $$ B_2\Bigl(x,\frac{r}{c}\Bigr)\subset B_1(x,r). $$ What does this say about the topologies? 3) Observe that $\delta$ depends on $x$. You have to choose a bounded neighborhood $U$ of $(x_0,y_0)$ and $M$ such that $|x|\le M$ ...


0

In its more general understanding, the Cauchy-Schwarz inequality compares the inner product ($\langle\ ,\ \rangle$) of two vectors of a space with their norm ($\|\ \|$) (this does require the space to have both an inner product and the corresponding norm). It states that in such a space $E$ : $ \forall (x,y)\in E^2\ \ \ |\langle x,y \rangle| \leq \| x\| ...


1

This is an answer from the geometrical view of inner product. Let $a = [a_1, a_2, \cdots, a_n]^T$ and $b = [b_1, b_2, \cdots, b_n]^T$. The inner product between $a$ and $b$, denoted as $\langle a, b \rangle$, is defined as $$ \langle a, b \rangle = \sum_{i=1}^{n} a_ib_i $$ Its geometrical reinterpretation shows that the following equation holds: $$ ...


2

It is precisely the application of Cauchy-Schwartz, which states: $$||x||\cdot||y|| \geq \langle x,y\rangle,$$ or, if you square this, you get $$||x||^2\cdot||y||^2 \geq \langle x,y\rangle^2.$$ Now, if the components of $x$ are $x_i$ and the components of $y$ are $y_i$, this equation becomes $$\left(\sum_{i=1}^n x_i^2\right)\cdot \left(\sum_{i=1}^n ...


0

Cauchy-Schwarz says that $$ (a_{i1}x_1+a_{i2}x_2+\cdots+a_{in}x_n)^2\leq (a_{i1}^2+a_{i2}^2+\cdots+a_{in}^2)(x_1^2+x_2^2+\cdots+x_n^2). $$ which is exactly that inequality you point to (note that this is just the inner sum we work on, so we do this for all $i$). If this was not what you asked for, please explain your problem more.


1

Since $\Vert \cdot \Vert$ is the induced norm, we have $\Vert v \vert = \sqrt{\langle v,v \rangle}$. If $v = 0$, then the inequality $$\Vert v \Vert \leq \sup \left\{ \langle v,w \rangle \, : \, w \in V, \Vert w \Vert = 1 \right\} $$ is trivial, so assume $0 \neq v \in V$. Note that for $w := \frac{v}{\Vert v \Vert}$ we have $\Vert w \Vert = 1$. So we ...


0

By Cauchy-Schartz inequality for $v,w \in V$ you have $\vert \langle v,w \rangle \vert \le \Vert v \Vert \Vert w \Vert$. Hence for $\Vert w \Vert =1$ you have $\vert \langle v,w \rangle \vert \le \Vert v \Vert$. Which proves that $\sup\{\vert \langle v,w \rangle \vert | \ w \in V, \ \Vert w \Vert =1\} \le \Vert w \Vert$ To prove the reverse inequality you ...


1

The point is that we want to show that we can get $\|x_n\|$ as close to $\|x\|$ as we wish based on the assumption that we can get $\|x_n - x\|$ as close to $0$ as we wish. The bound $|\|x_n\| - \|x\|| \leq \|x_n -x\|$ guarantees this, because we can just take the limit on both sides: $$\lim_{n \to \infty} |\|x_n\| - \|x\|| \leq \lim_{n \to \infty} \|x_n - ...


1

Yes, whenever you have a norm $\|\cdot\|$ on some space, it automatically gives you a metric $d(f,g) = \|f-g\|$. (One says that the norm induces the metric.) So, the norm $\|f\|=\int_0^1 |f(x)|\,dx$ induces the metric $d(f,g)=\int_0^1 |f(x)-g(x)|\,dx$ the norm $\|f\|=\sup_{0\le x\le1} |f(x)| $ induces the metric $d(f,g)=\sup_{0\le x\le 1}|f(x)-g(x)|$, ...


1

Your proof is all right except in (2), it should be $||y_1||_1+||y_2||_1$ rather than $||y_1||_1+||y_2||_2$.


1

About the triangle inequality for the $\max$ norm, it is better to take the $\max$ in the "right order": $$|y_1(x)+y_2(x)| \leq |y_1(x)| + |y_2(x)| \leq \|y_1\|_M+\|y_2\|_M.$$So $\|y_1\|_M+\|y_2\|_M$ is an upper bound for $\{ |y_1(x)+y_2(x)| \mid x \in [a,b]\}$, and now you take the $\max$ to obtain $$\|y_1+y_2\|_M \leq \|y_1\|_M+\|y_2\|_M.$$ About the ...


0

Whenever $f_n\rightarrow g$ you can change a value of $g$ such that $g$ is not continuous. But changing a value of $g$ does not change $\int_a^b |f_n(x)-g(x)|dx$ so you still have $f_n\rightarrow g$...


1

Hint: Consider the set of functions $$f_n(x) = \begin{cases} 0, & x \in \left[0, \frac{1}{2}-\frac{1}{2n}\right) \\ nx + \frac{1-n}{2}, & x \in \left[\frac{1}{2}-\frac{1}{2n},\frac{1}{2}+\frac{1}{2n}\right] \\ 1, & x\in\left(\frac{1}{2}+\frac{1}{2n},1\right]\end{cases}$$ Does this seemingly converge to anything in the $L^1$ norm? If so, is that ...


0

The triangle equality is true in general for any inner product space, and the defining properties of inner product spaces are pretty easy to see in this case (it looks like you already have most of them). The proof of the triangle inequality goes through Cauchy-Schwarz, i.e. $|(f,g)| \leq ||f||\cdot||g||$. To see Cauchy-Schwarz, observe that by ...


3

Let $ \Bbb{F} $ denote the base field of $ V $. If we assume that $ \star $ is a binary operation on $ V $ that turns $ V $ into an $ \Bbb{F} $-algebra, i.e., Left distributivity: $ x \star (y + z) = x \star y + x \star z $ for every $ x,y,z \in V $, Right distributivity: $ (x + y) \star z = x \star z + y \star z $ for every $ x,y,z \in V $, and ...


1

Note that for any $x\in \mathbb{K}^n$ $$ \|T(x)\|_1 = \sum_{i=1}^m |T(x)_i| \leq \sum_{i=1}^m\sum_{j=1}^n |a_{i,j}||x_j| \qquad (\ast) $$ For $1\leq j\leq n$, set $$ \alpha_j = \sum_{i=1}^m |a_{i,j}| $$ Then rearrange $(\ast)$ to get $$ \|T(x)\|_1 \leq \sum_{j=1}^n \alpha_j |x_j| \leq \|A\|_1 \|x\|_1 $$ This proves that $\|T\| \leq \|A\|_1$. For the reverse ...


1

If $g(x_0) = 0$, then for $\epsilon > 0, \exists \delta > 0$ such that $$ |x-x_0| < \delta \Rightarrow |g(x)| < \epsilon $$ Choose $f \in E$ such that $\|f\|_{\infty} = 1$, and $$ f(x) = 0 \text{ if } |x-x_0| \geq \delta $$ Then $\|fg\|_{\infty} < \epsilon$. Hence, there is no $a>0$ such that $$ a\|f\|_{\infty} \leq \|fg\|_{\infty} ...


0

Let $M=(B-A\circ X)$, then $$\eqalign{ d\,\|M\|_1 &= {\rm sign}(M):dM \cr &= {\rm sign}(M):(-A\circ dX) \cr &= -{\rm sign}(M)\circ A:dX \cr \frac{\partial\,\|M\|_1}{\partial X} &= -A\circ{\rm sign}(M) \cr }$$ in the case of the entrywise (Manhattan) norm. And the sign function is applied entrywise. If instead you meant the Schatten ...


2

In the case $q<p^*$ the Rellich–Kondrachov theorem applies, if your domain $\Omega$ has the required properties. If $q=p^*$, the $L^q$ norm is still controlled by the $W^{1,p}$ norm; and since a weakly convergent sequence is norm-bounded, $L^q$ norm is bounded. Moreover, you will have weak convergence in $L^q$, since linear maps preserve weak ...


0

Let $M=XA^T$, then taking the differential leads directly to the derivative $$\eqalign{ f &= \frac{1}{2}\,M:M \cr df &= M:dM \cr &= M:dX\,A^T \cr &= MA:dX \cr &= XA^TA:dX \cr \frac{\partial f}{\partial X} &= XA^TA \cr }$$ Your question asks for the {$i,j$}-th component of this derivative, which is obtained by taking its ...


0

See the method described in the answer to your other question here. In particular, take $x = (1,0)$ and $y = (0,1)$.


0

Pick two vectors $x$, $y$ in the space under consideration, which is presumably $\mathbb{R}^2$. Calculate $\|x+y\|$, $\|x-y\|$, $\|x\|$ and $\|y\|$ using the norm you are given, in this case $\|\cdot\|_1$. See whether the parallelogram rule holds for this particular $x$, $y$. If it fails, then you know that the norm does not come from an inner product.


1

First of all: since all norms on a finite-dimensional space are comparable, the condition could be simpler stated as $$ C_1\|x-y\| \leq d(x,y) \leq C_2\|x-y\| $$ where $\|\cdot \|$ is a norm of our choice, e.g., Euclidean. Second: the answer is negative, for example $$d(x,y) = |x-y|+\min(|x-y|,1)$$ is a translation-invariant metric on $\mathbb{R}$ that ...


1

Converted to an answer by request. Well, they didn't make a mistake, but the $2$-norm is not the $\infty$-norm, so they answered a different question. Both equations, $\|u−v\|_\infty=3$ and $\|u−v\|_2=\sqrt{13}$ are correct, though.


0

I don't think your "result so far" is entirely accurate. Defining $\mathbb I=$ (matrix of all ones), $\,E=({\mathbb I}-I)$, and using the Hadamard ($\circ$) product, we can use $(-E\circ X)$ to replace those bulky $({\rm diag}(X)-X)$ terms. Ignoring the L1-term, and using the Frobenius (:) product, the function to be minimized is $$ f = ...


1

Assuming $0\leq \lambda \leq 1$, $$||\lambda x_1 + (1-\lambda)x_2 - y|| = ||\lambda x_1 + (1-\lambda)x_2 - (\lambda + (1-\lambda))y||$$$$\leq ||\lambda x_1 - \lambda y|| + ||(1-\lambda)x_2 - (1-\lambda)y||$$$$ = \lambda||x_1 - y|| + (1-\lambda)||x_2 - y||$$$$ = \lambda||x_1 - y|| + (1-\lambda)||x_1 - y|| = ||x_1 - y||$$


0

Apply the triangle inequality to $ ||\lambda(x_1-y)+(1-\lambda)(x_2-y)||$.


0

Here I provide a solution only for an inner product space $H$. Let $a=x_1-y$ and $b=x_2-y$. Then $\|a\|=\|b\|$. WLOG, set $\|a\|=\|b\|=1$ and hence $(a,b)\le 1$. If $(a,b)=1$, then $a=b$. If $(a,b)<1$, then for $\lambda\in [0,1]$, \begin{eqnarray} f(\lambda)&=&\|\lambda x_1+(1-\lambda)x_2-y\|^2\\ &=&\|\lambda a+(1-\lambda)b\|^2\\ ...


0

This is false. Let $$ x_1=x_2=\begin{bmatrix}1&0\\0&1\end{bmatrix},\ \ y_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}\ \ y_2=\begin{bmatrix}0&0\\0&1\end{bmatrix}. $$ Then $\|x_1\|=\|x_2\|=\|y_1\|=\|y_2\|=1$, but $$ \left\|\begin{bmatrix}x_1& x_2\\0&0\end{bmatrix}\right\| = ...


0

Let $Y = \lVert X \rVert^2$. Then $Y/\sigma^2 \sim \chi^2_p$. And you wish to find $EY^{-1/2}$. This is $$ \frac{1}{\sigma}\cdot\dfrac{1}{2^{p/2}\Gamma(p/2)}\int_0^\infty y^{-1/2} y^{p/2 - 1} e^{-y/2}dy$$ $$ = \frac{1}{\sigma} \dfrac{2^{(p-1)/2}\Gamma((p-1)/2)}{2^{p/2}\Gamma(p/2)} = \frac{1}{\sigma\sqrt{2}}\dfrac{\Gamma((p-1)/2)}{\Gamma(p/2)}$$


0

The norm can be written as $$ \lVert y \rVert_2^2 = y^T y. $$ Hence, $$ \lVert A^T x \rVert_2^2 = (A^T x)^T A^T x = x^T A A^T x = x^T B x. \tag{1} $$ You also have $$ B^{-1} = (A A^T)^{-1} = (A^T )^{-1} A^{-1} = (A^{-1})^T A^{-1}, $$ and now you can apply (1) to $x^T B^{-1} x$.


1

One can write $$x^TBx=x^T(AA^T)x=(x^TA)A^Tx=(A^Tx)^TA^Tx.$$ Same thing for the second equality.Maybe the true question is why $(A^T)^T=A$ and $(A^{-1})^T=(A^T)^{-1}$ ?


0

Let $$ S = \{(y_n) \in \ell^{\infty} : \exists C > 0 \text{ such that } |y_n| \leq C/n\quad\forall n\in \mathbb{N}\} $$ If $(x_n) \in \ell^{\infty}$ then $T(x_n) \in S$ with $c = \|(x_n)\|_{\infty}$, and conversely, if $(y_n) \in S$, then the sequence $(x_n)$ defined by $$ x_n := ny_n $$ is in $\ell^{\infty}$ and satisfies $T(x_n) = (y_n)$. Hence, $R(T) = ...


0

Let us denote two norms as $\mu(x) = \|x\|_1 $ and $\nu(x) = \|x\|$. From the definition of $\|x\|_1 $ we have $\mu(x) = \nu(x) + |f(x)|$. Two norms $\mu$ and $\nu$ are equivalent if there exist two constants $c,C\in\mathbb{R}$ such that $\forall x\in X$ $c\nu(x) \leq \mu(x) \leq C\nu(x)$. First, since $\mu(x) = \nu(x) + |f(x)|$ and $|f(x)|>0$ we ...


1

The answer is AFFIRMATIVE First suppose that the two norms are equivalent and $a$ and $b$ be such that $a\|x\|\leq \|x\|_1\leq b\|x\|$ for all $x\in X$. Then $|f(x)|=|\|x\|_1-\|x\||\leq \|x\|_1+\|x\|\leq (1+b)\|x\|$ for all $x\in X$. Therefore, $f$ is continuous. Conversely, let us assume that $f$ is continuous. It is clear from the definition of ...


1

It is clear that $\|x\| \le \|x\|_1$. If $f$ is continuous, it is a bounded operator and so there is some $M$ such that $|f(x)| \le M\|x\|$. Then $\|x\|_1 \le (1+M) \|x\|$, and so $\|\cdot\|_1$ and $\|\cdot\|$ are equivalent. If the two norms are equivalent, there is some $L$ such that $\|x\|_1 \le L \|x\|$, and so $|f(x)| \le (L-1) \|x\|$. Hence $f$ is ...


4

If $\{ x_{n} \}$ converges to $x$ in $\|\cdot\|_1$, then it converges in $\|\cdot\|$ because $$ \|x-x_n\| \le \|x-x_n\|_1. $$ Conversely, if $\{ x_n \}$ converges to $x$ in $\|\cdot\|$, then $\{f(x_n)\}$ converges to $f(x)$ because $f$ is continuous; therefore, $\{ x_{n} \}$ converges to $x$ in $\|\cdot\|_1$.


1

Hint. It suffices to prove the following: $$ \|x_n-x\|\to 0 \quad\text{iff}\quad \|x_n-x\|_1\to 0. $$


1

I think the inequality $|Ax| \leq \| A \| |x|$ really doesn't imply what you need. Here is another suggestion of how to see it: from the definition of $\|A\|,$ which says $$\|A\| = \sup_{|x| \leq 1} |Ax|,$$ it is clear that if $|Ax| \leq \lambda |x|$ for every $x \in \mathbb{R}^n,$ then the set of real numbers $\{|Ax|: |x| \leq 1\}$ is a subset of the set ...


1

If $|Ax|\leq \lambda|x|$ for all $x\in\mathbb{R}^n$, then in particular it holds for all $|x|\leq 1$. Then $$\|A\|=\sup_{|x|\leq 1}|Ax|\leq\sup_{|x|\leq 1}\lambda|x|\leq\lambda.$$


2

By definition $||A||$ is the supremum of all $|Ax|$ where $x$ ranges over the unit ball. Now assume that $|Ax| \leq \lambda|x|$ for all $x \in \mathbb{R}^n$. First observation is that $\lambda \geq 0$ since $|Ae_1| \geq 0$ and $|e_1| = 1$ (here $e_1$ is some vector with norm $1$). Let $x \in \mathbb{R}^n$ with $|x| \leq 1$. Then by definition $|Ax| \leq ...


1

This comes from an equivalent definition of the Norm $\| A \|$: $$ \| A \| := \text{inf} \{\lambda \geq 0 | \forall x \in \mathbb{R}^n: |Ax| \leq \lambda |x| \} $$


0

We look for the linear and the bilinear terms in the multivariate Taylor approximation at $x$ (in subscript): $$f(x+u) \approx f_x + Df_x\,(u) + \tfrac12 Hessf_x\,(u,u).$$ Let $f(x)=\|Ax\|^2$, then $$ \langle A(x+h),A(x+h)\rangle=\|Ax\|^2+2\langle Ax,h\rangle+\langle Ah,Ah\rangle $$ The Hessian term can be written as $\langle Ah,Ah\rangle=\tfrac12 \langle ...


2

Consider the subset consisting of the $e_n$'s. It's bounded, because each sequence has norm $1$. Since the distance between any two distinct sequences is $2$, any Cauchy sequence is constant and thus converges, so the subset is closed. However, it's not compact because, it does not have a convergent subsequence. The reason your sequence doesn't have a ...



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