# Tag Info

13

By triangle inequality \begin{align} \left\|ax + \left(1-a\right)y \right\|\le \left\|ax \right\| + \left\|\left(1-a\right)y \right\|= a\left\|x \right\| + \left(1-a\right)\left\|y\right\| \end{align}

6

Let $B_R=\{f\in K : \|\,f\|<R\}$. If $K$ were open, then $L(B_1)$ would be open, and as $0\in L(B_1)$, there would exist an $r>0$, such that $B_r\subset L(B_1)$. Observe that, if $f\in L(B_1)$, then $\,f(x)=xg(x)$, for some continuous $g$ with $\lvert g(x)\rvert <1$, for all $x$. Hence $$\lvert\, f(x)\rvert=\lvert xg(x)\rvert=x\lvert g(x)\rvert ... 6 Yes, this is true. First you should consider finitely many elements f_1, \dotsc, f_n \in I. Assume they have no common zero. Then the function f=\sum f_i\overline{f_i} has no zero, hence f is a unit. On the other hand f \in I, contradiction! To get a common zero for all elements of I, we have to - of course - use compactness: To find a suitable ... 6 The inner product is already determined by the norm using the Polarization identity so no need to try and build it in a non-constructive way. If you define a function \left< \cdot, \cdot \right> on X \times X by$$ \left< u, v \right> := \frac{||u + v||^2 + ||u - v||^2}{4} $$then clearly \left<u, u\right> = ||u||^2 for all u \in ... 5 If \lambda \ne 0, its kernel has codimension 1. Namely, if u\in X with \lambda(u) \ne 0, every member of X can be written as$$ x = \dfrac{\lambda(x)}{\lambda(u)} u + w$$where \lambda(w) = 0. 5 Let f,g \colon \mathbf N \to \mathbf N be given by$$ g(n) = n+1, $$and$$ f(n) = \begin{cases} n - 1 & n \ge 1\\ 27 & n = 0 .\end{cases} $$then neither f nor g is bijective, but f \circ g = \mathrm{id}_{\mathbf N} is. 5 Take$$ f(x)=\left\{\begin{array}{lll} x & \text{if} & x<\sqrt{2}, \\ -x & \text{if} & x\ge\sqrt{2}, \\ \end{array} \right. $$The f :\mathbb R\to\mathbb R  is continuous on every rational, but there is no way to make it continuous on \mathbb R! EDIT. If we knew that f was locally uniformly continuous, then the sought for g does ... 5 Ok, then the solution is given by the variation-of-constants-formula$$x(t)=e^{tA}x_0 +\int_0^t e^{(t-s)A}Bu_a(s)\ ds$$so your question amounts to asking whether$$\sup_{0\le t\le T}\left\|\int_0^t e^{(t-s)A}B(u_a-u_0)(s)\ ds \right\|$$(for any norm \|\cdot\| on \mathbb R^n) goes to 0 as a goes to 0. Now this is certainly the case, because this ... 4 This is an overkill I guess. Let \mu be a Radon measure on X, then \mu(K) <\infty. Also \| f_n\|_{L^1} is uniformly bounded by \mu(K) as |f_n|\le 1. By Lebesgue's dominated convergence theorem,$$\tag{1} \int_K f_n d\mu \to \int_X f d\mu.$$By Riesz' representation theorem, all bounded linear functional on C(K) is given by Radon measure, ... 4 You are sane. Your argument with the atomic decomposition is of course irrefutable, and f\in H^1. For an independent way to see this, we can use the fact that H^1(\mathbb R) functions can also be characterized as the real parts of the boundary values of holomorphic functions F on \mathbb C^+ with \sup_{y>0} \int_{-\infty}^{\infty} |F(x+iy)|\, dx ... 4 If x^2+y^2=1, then by Cauchy Schwartz, |\alpha x + \beta y| \leq (\alpha^2+\beta^2)^{1/2}. On the other hand, T\big(\frac{\alpha}{(\alpha^2+\beta^2)^{1/2}}+i\frac{\beta}{(\alpha^2+\beta^2)^{1/2}}\big)= (\alpha^2+\beta^2)^{1/2}. So, \|T\| = (\alpha^2+\beta^2)^{1/2}. 4 Let's assume 0 \le a < b. Define \Phi : L^2 [a,b] \to L^2[-b,b] by$$\Phi f(x) = \begin{cases} f(x) & \text{if } x \in [a,b] \\ 0 &\text{if } x \in [-a,a] \\ f(-x) & \text{if } x \in [-b,-a] \end{cases}$$The image of \Phi lies in the space of even functions L_E in L^2[-b,b]. Note that \{t, t^3, t^5, \cdots\} is orthogonal to ... 4 In the setting of normed linear spaces, a basis spans a dense subspace. To say the monomials constitute a basis of L^{2}[-1, 1] amounts to the assertion that for every function f in L^{2}[-1, 1], there exists a sequence (p_{n}) of polynomials such that$$ \int_{-1}^{1} |f - p_{n}|^{2} \to 0. $$The space of continuous functions on [-1, 1] is dense ... 4 Take an infinite-dimensional hilbert space with an orthonormal basis \{e_j\}, I'm sure you know that e_n \xrightarrow{w} 0... but, obviously,  \|e_n\| =1  Check your Hahn-Banach corollary again! 4 Let$$\sup_{n\in \mathbb N} \| f'_n\|_2 = D <\infty.$$Using the Fundamental Theorem of Calculus, if f = f_n,$$|f(x)| = |f(x) - f(0)| = \left|\int_0^x f'(s) ds \right| \le \sqrt x \|f'\|_2 \le D.$$So \{f_n\} has a uniform C^0 bound. Similarly,$$|f(x) - f(y)| \le \sqrt{|x-y|} \|f'\|_2 \le \sqrt{|x-y|}D.$$Thus the family \{f_n\} is ... 4 Let E\subseteq X. We set J_E:=\{f\in C(X):\text{f vanishes on a neighbourhood of E}\}. It is easy to see that J_E is an ideal in C(X). Take E=\{x\} for some x\in X. Then each function in I_{\{x\}} can be approximated uniformly by functions in J_{\{x\}}, but not every function such that f(x)=0 is zero on a neighbourhood of x. Thus ... 4 The derivative operator D=\frac d{dx} has eigenfunctions e^{cx} with eigenvalue c 4 noting the series expansion \log(1-x)=-\sum_{n=1}^\infty x^n/n, you see that the right-hand-side of your equation is just$$\log\frac{r_>}{L}-\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{r_<}{r_>}\right)^n\cos[n(\phi-\phi')]=\log\frac{r_>}{L}+{\rm Re}\,\log\left(1-\frac{r_<}{r_>} e^{i\phi-i\phi'}\right)={\rm ...

3

It is available on Gallica (the online platform of the France National Library) here.

3

The set $\{ u_0, v_1, u_1, v_2, u_2, \cdots \}$ is a complete orthonormal basis of $L^2(-\pi,\pi)$. So there is an isometric correspondence between $L^2(-\pi,\pi)$ and $\ell^2(\mathbb{N})$ defined by $$f \sim ((f,u_0),(f,v_1),(f,u_1),(f,v_2),(f,u_2),\cdots).$$ This is due to Parseval's equality. Using this correspondence, the operator $T$ looks ...

3

Hint: try checking that $P([0,1]) \ni p \mapsto p' \in P([0,1])$ is not bounded in the unit sphere. Here $P([0, 1])$ is the space of polynomials in $[0,1]$, with the sup norm.

3

For any sequence $y\in c_0$ we have $$\|x-y\|_\infty = \sup_n|x_n-y_n|\ge \limsup_{n\to\infty} |x_n-y_n|=\limsup_{n\to\infty} |x_n|$$ This gives a lower bound on the distance. To get the matching upper bound, let $y_n=x_n$ when $n\le N$, and $y_n=0$ otherwise. This is an element of $c_0$, and $$\|x-y\|_\infty = \sup_{n>N}|x_n| ... 3 The dual space of C(K) is the space of Radon measure \nu on K. For every bounded Borel function u on K, one can define \bar u \in C(K)^{**} by$$\bar u (\nu) = \int_K u d\nu,\ \ \ \ \forall \nu \in C(K)^*. $$Note that the canonical embedding \Phi: C(K) \to C(K)^{**} is given by$$\Phi f(\nu) = \nu(f) = \int_K fd\nu,$$thus if \Phi is ... 3 You are dealing with the convolution of f with$$ \phi_n(x) = \frac{1}{n}\chi_{[-n,0]}(x). $$That is,$$ (f\star\phi_n)(x)=\int_{-\infty}^{\infty}f(t)\phi_n(x-t)dt=\frac{1}{n}\int_{x}^{x+n}f(t)dt = f_n(x) $$Therefore \|f_n\|_1 \le \|f\|_1\|\phi_n\|_1 = \|f\|_1. If g \in \mathcal{C}_c^{\infty}(\mathbb{R}) (i.e., ... 3 We will show that \sigma(A) = \{0\}, and that 0 belongs to the residual spectrum. As you have shown, 0 is not an eigenvalue, and as \operatorname{im} A \subseteq \{x \in C[0,1]: x(0) = 0\}, A does not have dense image. Hence 0 \in \sigma_r(A). To see that A - \lambda is invertible for \lambda \ne 0, let y \in C[0,1] be given. We have to ... 3 Let ||x_n||<A Take M to be the subspace of finite linear combination of basis elements. Then it is easy to see that M is dense in H. \langle x_n,e\rangle \rightarrow 0 \forall e\implies \langle x_n,m\rangle \rightarrow 0 \forall m\in M Now fix y\in H. Take \epsilon >0 Choose m\in M such that ||m-y||<\epsilon A^{-1}  As ... 3 Let V := \langle E\rangle  be the span of E , i.e. the set of finite linear combinations of the orthonormal basis. It is easy to see \langle x_n, v\rangle \to 0. Now, let y \in H  be arbitrary and \epsilon >0. There is v \in V  with \Vert v -y\Vert <\epsilon , since V  is dense because E  is an orthonormal basis. Now,$$ |\langle ...

3

Theorem: Suppose $A : \mathcal{D}(A)\subseteq X\rightarrow X$ is a surjective linear operator on a complex Hilbert space $X$ for which $(Ax,x) \ge \|x\|^{2}$ holds for all $x\in\mathcal{D}(A)$. Then $A$ is densely-defined and selfadjoint. Proof: To show that $A$ is densely-defined suppose otherwise. Then there exists $y \perp \mathcal{D}(A)$. However, ...

3

This isn't true. For instance, if $X=\omega_1$ is the first uncountable ordinal, then $C_c(X)=C_0(X)$ is complete (since every continuous map $X\to \mathbb{R}$ is eventually constant), but $X$ is not compact.

3

The claim that $T$ is surjective implies range of $T^*$ being closed is not true. Take $X=c_{00}=Y$ the space of sequences with finite length. The dual space can be identified with $l^1$. Define $$Tx = (x_1, x_2/2, \dots, x_n/n,\dots).$$ Clearly, $T:X\to Y$ is injective and surjective, however $T^{-1}$ is not bounded. Let $g\in l^1$ be given. Then for ...

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