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## Hot answers tagged norm

4

$||(1,0)+(0,1)|| = ||(1,1)|| = 2^{1/p} \geq 2 = 1+1 = ||(1,0)||+||(0,1)||$ where in the critical step we use the fact that p is less than one. So we have a counterexample to the triangle inequality. Basically, the main reason these are not norms is that the unit ball is not convex, which means you can pick two points in the unit ball, like (0,1) and (1,0), ...

4

Let $x$ be a vector in $\mathbb R^n$ with non-negative entries. Assume $x_1$ is the largest of them. Then we have to prove $$x_1 \cdot \sum_{i=1}^n x_i \le c \sum_{i=1}^n x_i^2.$$ My idea is to estimate for $i\ne 1$ with $a>0$ $$x_1 x_i \le \frac1{2a}x_1^2 + \frac a2 x_i^2.$$ In order to obtain a balanced estimate, $$1+(n-1)\frac1{2a} = \frac a2$$ ...

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 $\|\cdot\|_1$.

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 $\| A^T A \| \| x \| \neq 0$.

3

As you know, $$1 + \|y\|_1 \le 1 + \sqrt{n-1}\|y\|_2$$ so it would suffice to show $$1 + \sqrt{n-1}\,\|y\|_2 \le \frac{1+\sqrt n}{2}(1+\|y\|_2^2)$$ In this inequality (if it's true), the fact that $\|y\|_2$ is the norm of a vector doesn't seem to have any significance... it's just a number, here. Since you requested only a hint, I'll stop here.

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 ...

3

$1.$ $T$ is surjective because of the Fundamental Theorem of Calculus. The reason that $T^{-1}$ does not exist is that $T$ is failing to be injective. $T$ is not injective because any two functions that differ by a constant will get mapped to the same function, e.g., $x$ and $x+c$, $c\in \mathbb{R}$ both get mapped to the constant function $1$. $2.$ Since ...

2

You essentialy have a inner product there: $\langle x,y \rangle_A = x^TAy$. Bilinearity is obvious. Symmetry comes from the fact that $A = A^T$. And $\langle x,x \rangle_A \geq 0$ and $\langle x,x\rangle_A = 0 \iff x = 0$ comes from positive-definiteness. Once you convince yourself of that, it is done: we have Cauchy-Schwarz's inequality, and with this you ...

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 ...

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 ...

2

$1.$ If $T$ is bounded then it is not hard to see that $T$ maps bounded sets to bounded sets. Conversely, let us assume that $T$ maps bounded sets to bounded sets. In particular $T(B_X)$ will be bounded, where $B_X$ is the closed unit ball of $X$. So, there exists $c>0$ such that $\|Tx\|\leq c\|x\|$ for all $x\in B_X$. Now, let $y\in X$ be any arbitrary ...

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 $$\left(\sum_{i=1}^n x_i^2\right)\cdot \left(\sum_{i=1}^n ... 2 Consider the subset consisting of the e_n's. It's bounded, because each sequence has norm 1. Since the distance between any two distinct sequences is 2, any Cauchy sequence is constant and thus converges, so the subset is closed. However, it's not compact because, it does not have a convergent subsequence. The reason your sequence doesn't have a ... 2 One way of playing with this is to note that d'(x,y) = d(f(x),f(y)) where f(t) = t/(1+|t|) maps \mathbb{R}\to (-1,1). In other words, d' is the standard metric on the interval (-1,1) pulled back by the homeomorphism f. So this problem boils down to showing that f is continuous and continuously invertible. 2 Show that an open ball wrt d is an open ball wrt d' (not necessarily of the same radius), and vice versa. Then, since the open balls form a basis for the topology, the topologies are the same. 2 (a) T being continuous has nothing to do with T mapping continuous functions. T\colon C^0 \to C^0 is continuous iff it is bounded, which means that the number \def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|}$$ \norm T := \sup_{\abs f \le 1} \abs{Tf} $$(the norm of T) is finite. Let f \in C^0 with \abs f \le 1. Then we have ... 2 By Fundamental Theorem of Calculus you can see that the range of T is R(T) =\{f\in C^1[0,1]\mid f(0)=0\} where C^1[0,1] is the space of all continuously differentiable functions. T^{-1} is linear is easy to see. But T^{-1} is not bounded. To see this first of all notice that T^{-1}(x(t)) = x'(t), i.e., T^{-1} is the differentiation operator ... 2 If B=[b_1,\ldots,b_n] is the column partitioning of B, then the definition of the Frobenius norm and consistency of the matrix and vector 2-norms gives$$ \|AB\|_F^2=\sum_{i=1}^n\|Ab_i\|_2^2\leq\|A\|_2^2\sum_{i=1}^n\|b_i\|_2^2=\|A\|_2^2\|B\|_F^2. $$2 By definition ||A|| is the supremum of all |Ax| where x ranges over the unit ball. Now assume that |Ax| \leq \lambda|x| for all x \in \mathbb{R}^n. First observation is that \lambda \geq 0 since |Ae_1| \geq 0 and |e_1| = 1 (here e_1 is some vector with norm 1). Let x \in \mathbb{R}^n with |x| \leq 1. Then by definition |Ax| \leq ... 1 I think the inequality |Ax| \leq \| A \| |x| really doesn't imply what you need. Here is another suggestion of how to see it: from the definition of \|A\|, which says$$\|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.$$1 This comes from an equivalent definition of the Norm \| A \|:$$ \| A \| := \text{inf} \{\lambda \geq 0 | \forall x \in \mathbb{R}^n: |Ax| \leq \lambda |x| \} $$1 Your proof is all right except in (2), it should be ||y_1||_1+||y_2||_1 rather than ||y_1||_1+||y_2||_2. 1 About the triangle inequality for the \max norm, it is better to take the \max in the "right order":$$|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.$$About the ... 1 An "abstract" construction can be done as follows: Take u the solution of the problem -\Delta u= 1 in \Omega and u=0 on \partial \Omega. If \Omega is smooth enough then u\in H^2(\Omega) by elliptic regularity. On the other hand, each partial derivative is harmonic so if u\in H_0^2(\Omega), this would mean that the partial derivatives have ... 1 The answer is AFFIRMATIVE First suppose that the two norms are equivalent and a and b be such that a\|x\|\leq \|x\|_1\leq b\|x\| for all x\in X. Then |f(x)|=|\|x\|_1-\|x\||\leq \|x\|_1+\|x\|\leq (1+b)\|x\| for all x\in X. Therefore, f is continuous. Conversely, let us assume that f is continuous. It is clear from the definition of ... 1 It is clear that \|x\| \le \|x\|_1. If f is continuous, it is a bounded operator and so there is some M such that |f(x)| \le M\|x\|. Then \|x\|_1 \le (1+M) \|x\|, and so \|\cdot\|_1 and \|\cdot\| are equivalent. If the two norms are equivalent, there is some L such that \|x\|_1 \le L \|x\|, and so |f(x)| \le (L-1) \|x\|. Hence f is ... 1 Hint. It suffices to prove the following:$$ \|x_n-x\|\to 0 \quad\text{iff}\quad \|x_n-x\|_1\to 0. $$1 Hint: Consider the set of functions$$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 ...

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.

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