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1

For $d_1$ consider $f_n(x) = [\sin(nx)]/n, n =1,2,\dots$ Then $f_n \to 0$ in the $1$ norm, but $d_1(f_n) = 1$ for all $n.$

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For $d_1$ you can refer to the comment, and even try to find a counter-example by reading the proof for $d_2$ and see where the proof can't be applied for $d_1$ and find a counter-example from here. For the norm $||.||_2$, you can show the continuity by noticing that : $$|d_2(f)-d_2(g)|=|f'(0)-g'(0)|\le ||f'-g'||_{\infty} \le ||f'-g'||_{\infty} + ... 1 Let z be such that |z^* x| = \|z\|' \|x\| = \|z\|' = \sup_{v:\|v\|=1} |z^* v| as given by the lemma. By re-scaling z, we may assume |z^* x|=1 Then with B=yz^*, we have \|B\| = \sup_{v : \|v\|=1} \|yz^* v\| = \|y\| \sup_{v : \|v\|=1} |z^* v| = \|y\| |z^* x| = 1. Finally, Bx = y(z^*x). It is not clear to me how we can show that z^* x=1; ... 0 Let \varepsilon be smaller than \min\{b-a,1\} and f(x)=g(x):=\mathbf 1_{[a,a+\varepsilon]}(x). Then$$\lVert fg\rVert_p=\left(\int_{[a,b]}\mathbf 1_{[a,a+\varepsilon]}(x)\right)^{1/p}=\varepsilon^{1/p}$$and$$\lVert f\rVert_p\lVert g\rVert_p=\varepsilon^{2/p}.$$1 If your "length" function is actually a quadratic form, you can use the notation:$$q(x) = a_{11}x_1^2 + a_{12}x_1x_2 + \cdots + a_{nn}x_n^2$$If your "norm" however is subadditive and separates points, but the only thing it doesn't have is the "absolute" part of absolute homogeneity, then I don't think there's a word for that concept (but don't quote me ... 4 The answer is yes. Let$$J = \left( {\begin{array}{*{20}c} \hfill {J_1} & \hfill {} & \hfill {} & \hfill {} \\ \hfill {} & \hfill {J_2} & \hfill {} & \hfill {} \\ \hfill {} & \hfill {} & \hfill \ddots & \hfill {} \\ \hfill {} & \hfill {} & \hfill {} & \hfill {J_n} \\ \end{array}} \right)$$be ... 1$$||u-v||^2=<u-v,u-v>$$By linearity of the scalar product :$$||u-v||^2=<u,u>-<u,v>-<v,u>+<v,v>$$By symetry of the scalar product :$$||u-v||^2=<u,u>-2<u,v>+<v,v>=||u||^2+||v||^2-2<u,v>$$Do the same with the other term of the left part of the equality, and you can conclude. 1 Take some f \in C[0,1]. Set$$g(x) := \max(f(x),0),$$then clearly g \in B. To see that this is the closest element in B, look at$$ \|f-g\|_2 = \int_{0}^{1} |f(x) - g(x)|^2 dx = \int_{[f\geq 0]} |f(x)-g(x)|^2 + \int_{[f<0]} |f(x)-g(x)|^2$$Obviously, on the set [f\geq 0] we have f=g such that the first integral vanishes. But on set where f is ... 2 You are right. And it has to be that way, because the norm that the Hermitian product induces on \mathbb C^n is the same as the Euclidean norm on \mathbb R^{2n}. Therefore, for a fixed x\in\mathbb C^n, the set$$ \{ y\in\mathbb C^n \mid \|x+y\|^2 = \|x\|^2+\|y\|^2 \} has dimension 2n-1 over \mathbb R -- and therefore it cannot be a linear ... 2 This is not possible For simplicity, just take r=2 and the nonzero singular values of C are 1,1. So that the F norm of C is \sqrt2. Now you may put the nonzero singular values of A as 1/N, N and those of B as N, 1/N aligned accordingly Then A,B have the same F-norm \sqrt{N^2+1/N^2}, but N can be as big as you like 0 Actually there is an answer, but somewhat complex. Denote the SVD decomposition of X by USV', denote qr(DU) = QR, and denote SVD decomposition of RS by U_1S_1V_1' then the SVD decomposition of DX is (QU_1)S_1(VV_1)'. Proof: DX= DUSV'=QRSV'=(QU_1)S_1(V_1'V') = (QU_1)S_1(VV_1)'. QU_1 is unitary and VV_1 is unitary and S_1 is diagonal. ... 1 Assuming your * is matrix multiplicationJ = \pmatrix{4cx+1 & 0\cr 0 & 4cy+1\cr}$$so this will be a contraction with the Euclidean norm in any convex region where |4cx+1|<1 and |4cy+1|<1. For this to be true in your rectangle (x,y) \in [0.93, 1.52] \times [0.41, 1] you need -25/76 < c < 0. 3 So this is my final answer for any interested party answer and thank you @joy \|\frac{x+y}{2}\|^2 \leq \|\frac{x+y}{2}\|^2 + \|\frac{x-y}{2}\|^2 = \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2 and since x \neq y: \|\frac{x+y}{2}\|^2 < \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2=\frac{1}{2}r^2+\frac{1}{2}r^2=r^2 Hence: \|\frac{x+y}{2}\| < r 4 Use paralleogram law for \frac{x}{2} and \frac{y}{2} to obtain ||\frac{x+y}{2}||^2 + ||\frac{x-y}{2}||^2 = \frac{1}{2}||x||^2 + \frac{1}{2}||y||^2 and so you get 1 in the right hand side. Since the LHS is a sum of two non-negative terms, you get the desired inequality since x\neq y . 1 At first I thought one or the other inequality must be obvious, but I don't see it after a little thought. Big Hint: It's trivial from the Closed Graph Theorem. -1 For a complex variable a, |a|^2 derivative exist only at zero. Try evaluating at x_0+i*y_0 not zero. Approach from y=y_0 and you get 2*x_0 but approach from x=x_0 and you get -2*y_0*i. And at zero you get limit r^2 over re^(thetai) that go to zero. -1 For a complex variable a, |a|^2 derivative exist only at zero. Try evaluating at x_0+i*y_0 not zero. Approach from y=y_0 and you get 2*x_0 but approach from x=x_0 and you get -2*y_0*i. And at zero you get limit r^2 over re^(thetai) that go to zero. 0 In Rudin's PMA (Theorem 6.27) \gamma is assumed continuously differentiable. The proof given there works for general normed spaces with no changes: see below. However, it invokes the integral triangle inequality$$ \left\|\int_a^b \mathbf f(t)\,dt \right\| \le \int_a^b \left\|\mathbf f(t)\right\|\,dt \tag{1}$$with \mathbf f = \gamma'. Since the ... 1 In fact, we can show more - that every matrix A which preserves the lengths of vectors in \mathbb{R}^2 is a rotation or a reflection. 1 No, you can't obtain this norm from an inner product. For instance, the parallelogram law doesn't hold for X=(1,0) and Y=(0,1):$$ \|X+Y\|^2+\|X-Y\|^2=(|1|+|1|)^2+(|1|+|-\!1|)^2=8\ne2\|X\|^2+2\|Y\|^2\;. $$2 Sure, for example, take  A  to be a rotation matrix, in other words any matrix of the form$$ A = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} $$We can check directly that the norm is conserved. Let  x = (a, b)^T , then we have$$ Ax = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & ...

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Yes. Pick any matrix with orthonormal column vectors.

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The first matrix norm is called the Frobenius norm, it's natural as in that it's the default Euclidean norm if the matrix were interpreted as a vector in $\mathbb{R}^{m \times n}$. The second matrix norm for $A \in \mathbb{R}^{m\times n}$ is the operator norm given by the linear operator $x \mapsto Ax$, it is naturally induced by the norms you choose for ...

1

Functional analysis has a lot to do with spaces of functions, as its name suggests (more so historically, but it still does). Using single bars for the norm of a function is ambiguous because $|f|$ also means the function $|f|(t) = |f(t)|$. Double bars $\|f\|$ eliminate the ambiguity. By the way, in some function spaces it is important to observe that ...

2

Hint: Show that if $E$ is a vector space finite dimensional then all norms in $E$ are equivalents, i.e., if $\|\cdot\|_1$ and $\|\cdot\|_2$ are norms in $E$ then there are $c_1,c_2 >0$ constants such that $$c_1 \|x\|_1\le\|x\|_2\le c_2\|x\|_1$$ for every $x\in E.$

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The trick is to first observe that $$||x||_{\infty}^p=\sup_n|x(n)|^p\leq \sum_{n}|x(n)|^p=||x||_p^p$$ hence $||x||_{\infty}\leq ||x||_p$. If $p<q<\infty$, then $$||x||_q^q=\sum_n|x(n)|^q\leq ||x||_{\infty}^{q-p}\sum_n|x(n)|^p=||x||_{\infty}^{q-p}||x||_p^p\leq ||x||_p^q$$ with the last inequality using the $\infty$ case. Taking $q$th roots shows that ...

1

First at all $c$ is not equal to $d$. If $x_n\to x$ under $||\cdot||_1$ then for all $\epsilon>0$ there is a $N$ such that for all $n>N$ we have $||x_n-x||<\epsilon$. Now from $$c\|x_n-x\|_2 \leq \|x_n-x ||_1 \leq d\|x_n-x\|_2$$ it is clear that $$\|x_n-x\|_2 \le \frac{\epsilon}{c}$$ One can deduce that $x_n\to x$ under $\|\cdot\|_2$ and ...

1

In a finite dimensional real (or complex) vector space all norms are equivalent. That means (following your notation) there are $c,d\in \mathbb R$ such that $$c||x||\leq ||x||_2\leq d||x||$$ Now the properties you want to show follows from this inequality.

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The statement is true. More generally, for every $\alpha\in (0,1)$ the metric space $(\mathbb{R}^n, \|x-y\|^\alpha)$ can be isometrically embedded into a Hilbert space (you only need $\alpha=1/2$ here). This is Corollary 4.8 in the book Embeddings and extensions in analysis by Wells and Williams, who attribute this result to John von Neumann. The ...

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First, notice that if we have two vectors $$v = \begin{pmatrix} v_1 \\ v_2 \\v_3 \end{pmatrix} \ \ w = \begin{pmatrix} w_1 \\ w_2 \\w_3 \end{pmatrix}$$ we can rewrite the dot product $$\langle v, w\rangle = v^H \times w = \begin {pmatrix} v_1^* & v_2^* & v_3^*\end{pmatrix} \times \begin{pmatrix} w_1 \\ w_2 \\ w_3 \end{pmatrix}$$ Now, we wish ...

3

Note that if $\|x\| < 1$ and $z^T x \ge 0$, then $z^T {x \over \|x\|} \ge z^T x$. Hence $\sup \{z^T x : \|x \| \le 1 \} = \sup \{z^T x : \|x \|= 1 \}$.

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You're correct. But note that in the BV definition, there is no benefit in taking $\|x\|<1$, so the definition may as well have stipulated that $\|x\|=1$.

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In general: continuity can be defined only for functions between topological spaces. It seems that you refer to a definition of continuity in terms of limits, and this is usually done in a metric space (that is also a topological space) (see here) . A vector space of matrices as, for example, $M(n, \mathbb{R})$ (that is a vector space) becomes a metric ...

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Well you need a notion of norm in order to speak of convergence,i.e. $$A_jB_j\longrightarrow AB\mathrm{\ \ iff\ \ }\forall\epsilon>0,\exists N>0\mathrm{\ s.t.\ }j>N\Rightarrow\Vert AB-A_jB_j\Vert<\epsilon$$ Anyway here you don't need to go back to the definition to prove that the product is continuous. You basically need properties which are ...

0

From your link: see the paragraph starting with Notation: on page 3.

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If a linear map preserves right angles then it preserves all angles and we can then deduce that all it has to be a multiple of an orthogonal matrix, ie. $AA^T=1$. See this: ALL Orthogonality preserving linear maps from $\mathbb R^n$ to $\mathbb R^n$?

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Suppose $x=0$. Then, it is easy to show that $X=\{0\}$. Therefore $\mathrm{dim}X=0$. Suppose $x\neq 0$. Assume WLG that $\|x\|=1$. Then, there is at least one vector $u\in X$, $u\neq x$. Take $u^\prime=u-\langle u,x\rangle x$. It is easy to show that $0=\langle u^\prime,x\rangle$. Thus, $u^\prime=0$. Therefore $u-\langle u,x\rangle x=0$. That is ...

1

Pick $a_k = \frac{1}{2^k}$. Then $$b_n = \frac{1}{n}\sum_{k=1}^n a_k = \frac{1}{n}\sum_{k=1}^n \frac{1}{2^k} = \frac{1}{n}\bigg(1 - \frac{1}{2^{n+1}}\bigg) > \frac{1}{2n},$$ so $\sum b_n$ diverges

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It fails for the sequence $1,0,0,\dots$

0

Write $\|x\|_\infty = \max |x_i|$. Clearly $\|x\|_\infty \le \|x\|_p$ for any $p$. Then for $p < q$ you have $$\|x\|_q^q = \sum_{k=1}^n |x_i|^q \le \|x\|_\infty^{q-p} \sum_{k=1}^n |x_i|^p \le \|x\|_\infty^{q-p} \|x||_p^p \le \|x\|_p^q.$$ Thus $\|x\|_q \le \|x\|_p$.

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Yes, the dual-norm is that one! Furthermore, you should consider following: First: The embedding $$H^1(\Omega)\hookrightarrow L^2(\Omega).$$ This is because $$\| \cdot \|_{_{H^1(\Omega)}}^{^2} := \| D^1\cdot \|_{_{L^2(\Omega)}}^{^2} +\| \cdot \|_{_{L^2(\Omega)}}^{^2}$$ is a "stronger" norm than $\|\cdot\|_{_{L²(\Omega)}}$. Second: it follows from the ...

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As stated, this is false, since even $A$ itself could be singular under the given assumptions. To conclude that $A-B$ is nonsingular when $B$ is "small", one needs: $A$ to be nonsingular $\|B\| < \|A^{-1}\|^{-1}$ The reason this works is that $\|A^{-1}\|^{-1} = \min_{\|x\|=1}\|Ax\|$. Hence, assumption 2 implies $\|Bx\|<\|Ax\|$ for all $x$, making ...

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The first inequality is clear by the definition of the matrix norm. The second is clear by the definition and $\frac{|ABx|}{|x|}=\frac{|ABx|}{|Bx|}\frac{|Bx|}{|x|}$.

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The first inequality is immediate, since $$\|A\| = \sup_{x \in \mathbb{R}^n \setminus \{0\}} \frac{|Ax|}{|x|}\ge \frac{|Ax|}{|x|}.$$ For the second note that for $x\ne0$ with $Bx\ne0$, we have $$\|AB\| = \sup_{x\ne0} \frac{|ABx|}{|x|}=\sup_{Bx\ne0}\left( ... 2 I assume you want this \forall K>0, not all t. Since f(t)^Tf(t)\geq0, you can take \beta=(\int_0^{\infty}f(t)^Tf(t)\,dt)^{1/2}. 1 Just note that \|f\|_K^2=\int_0^K f(t)^Tf(t) \, dt\leq \int_0^\infty f(t)^Tf(t) \, dt for all K, since f(t)^Tf(t) is always nonnegative. 1 Note that$$ \left(\frac{x}{e^x}\right)'=\frac{e^x(1-x)}{e^{2x}} $$vanishes at x=1 (which is a maximum of the function). Hence,$$ p(f)=\frac1e\sup_{\|v\|=1}f(v)=\frac1e\|f\|^*. $$0 It is not. In general, for a_i \in \mathbb{C}, we have$$ |a_1+a_1+\cdots+a_n|^2\neq|a_1|^2+|a_2|^2+\cdots+|a_n|^2. $$6 Suppose A and B are positive operators on a finite-dimensional inner product space V and \|A + B\| = \|A\| + \|B\|. Because A and B are positive, A+B is also a positive operator. Thus there exists x \in V such that$$\|x\| = 1 \quad \text{and} \quad\|A+B\| = \langle (A+B)x, x \rangle.  Now \begin{align*} \|A+B\| &= \langle (A+B)x, x ...

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Let $v$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, i.e. $Av=\lambda v$. Then $(\lambda I-B)^{-1}(A-B)v=(\lambda I-B)^{-1}(\lambda I -B)v=v$. Hence by definition of the operator norm, $\left\|(\lambda I-B)^{-1}(A-B)\right\|\geq 1$. Question: Where did we use that $\lambda$ is not an eigenvalue for $B$?

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