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Hint: Fix $\delta >0$. You need to show that for each $f \in C^2[0,1]$, there is a function $g$ with $g(0)=g(1)$ such that $\|g-f\|<\delta$. Use the fact that continuous functions on compact intervals are bounded. Cut off your function $f$ at both ends sufficiently close that you won't affect the norm much, and then stitch it back up linearly so that ...

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I may answer my question by myself because I find a nice answer in Ringstrom's book "The Cauchy problem in general relativity". In Chapter 4, he shows me a energy: $$E_k=\frac{1}{2}\sum_{|\alpha\le k|}\int_{\mathbb{R}^n}[(\partial^\alpha\partial_t u)^2+|\nabla \partial^\alpha u|^2]dx$$ For the linear case, of course, this energy is conserved, so according ...

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Many have provided a good answer in the comments already, but I'll just spell out every detail here: It is a standard fact (in topology) that the image of a compact set under a continuous map is compact, and in $\mathbb{R}^n$, compact is equivalent to closed and bounded. So, for the continuous map $f:S\to \mathbb{R}$, $f(S)$ is a closed and bounded set in ...

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The second function space is a proper subset of the one, and they are both proper subsets of the set of Borel functions.

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Use that $k$ is uniformly continuous. Given $\epsilon>0$ there exists $\delta>0$ such that $$|c-y|<\delta\implies |k(c,s)-k(y,s)|<\epsilon\quad\forall s\in[\,0,1\,].$$

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Proving the existence of Schauder bases sometimes relies on the axiom of choice, but many important normed spaces have explicit Schauder bases. Suppose we know a (countable) Schauder basis for $X$, call it $\{x_i\}$. The subspace $X_0$ for which $\{x_i\}$ is a Hamel basis is incomplete. Explicitly ...

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Assume that $\lvert M_{ij}(t)\rvert \le \sqrt{C}$ and $\lvert N_{ij}(t)\rvert\le \sqrt{C}$. We have that $$\begin{split} \lvert (Mu)\cdot (Nu)\rvert &= \left\lvert \sum_{ijk} M_{ij}(t) v_j N_{ik}(t)u_k\right\rvert\\ &\le C\sum_{ijk} \lvert v_j u_k\rvert \\ &= C n \sum_{jk} \lvert v_j u_k\rvert \\ &= C n \sum_{j}\lvert v_j\rvert ... 2 You can always try e^{f(t)}, for an f of your choice. The exponential will take care of the logarithm, and you can design f to handle the root in whichever way you want. 1 As you said, some of the x_n might be zero: note that all x_n belong to the image of T^*, which could perfectly be finite-dimensional. For your second question, S^*Se_n=e_n, as S^*S is the projection onto the image of |T|. 1 First, the only continuous additive functions from \mathbb{R} to \mathbb{R} are of the form x\mapsto \alpha x, for some \alpha \in \mathbb{R}. To see this, let \alpha=f(1) and note that f(\frac{a}{b})=\frac{a}{b} \alpha, for all rationals \frac{a}{b}. Hence, by continuity, for all reals f(x)=\alpha x An additive map from \mathbb{R}\to ... 0 This is stated as an open problem in the 2002 survey Banach spaces of continuous functions on compact spaces by Godefroy. In the more recent survey Smoothness in Banach spaces. Selected problems by Fabian, Montesinos and Zizler (page 119) the weaker form of this problem is stated (also open): "Assume that X admits a C^\infty smooth norm. Does X ... 2$$\int_0^1 dt \int_{u(t)}^{u(t)+w(t)} dv f(t,v) =\int_0^1 dt \int_{0}^{1} (w(t)d\theta) f(t,u(t) +\theta w(t))\\ =\int_{0}^{1} d\theta \int_0^1 w(t)dt f(t,u(t) +\theta w(t)) $$Now using the Mean value theorem for integrals:$$ \exists \theta_0\in(0,1) \int_0^1 dt \int_{u(t)}^{u(t)+w(t)} dv f(t,v) =\int_0^1 w(t)dt f(t,u(t) +\theta_0 w(t)) $$0 We recall the following characterization of closure via sequences: Let C\subseteq X. The following are equivalent: i) x\in C^{-} ii) \exists \,(x_n)_{n\in\mathbb{N}}\subseteq C such that x_n\longrightarrow x Now, let \beta\in K, \beta x\in\beta B, so x\in B. As B is closed there is a sequence ... 2 This "result" is false. The rule of thumb is that spaces characterized by a supremum condition, such as$$\sup_{0<h<T} \frac{1}{h} \int_0^{T-h}\lVert u(t+h)-u(t) \rVert_{L^2}^2 <\infty \tag{1}$$are non-separable spaces of Lipschitz/Hölder type, in which smooth functions are not norm-dense. This is quite unlike H^1. As a concrete ... 3 The proof is reasonable. When working with Sobolev spaces, one should keep in mind that the elements are equivalence classes of functions (agreeing except on a null set). However, in W^{k,p} with kp>n we have a canonical representative of each equivalence class - namely, a continuous function. So it is understood, often without saying, that this ... 0 You can read about such duality in the first pages of Chapter 6 in Paulsen. Given \phi\in\text{UCP}(A,M_n(\mathbb C)), it can be identified with the state s_\phi on M_n(A) by$$ s_\phi([a_{kj}])=\frac1n\,\sum_{k,j}\phi(a_{kj})_{kj}. $$The hypothesis that \phi is zero on the compacts implies that so is s_\phi. So we can apply Glimm's Lemma to ... 2 The metric describes pointwise convergence on the dense set \lbrace f_n:n\in\mathbb N\rbrace which is a vector space topology (even locally convex) which is clearly coarser than the usual operator norm topology. Therefore, the metric cannot be complete because otherwise the open mapping theorem (for Frechet spaces) implies that the metric and the norm give ... 1 Converse fails in B(H) (Problem 102 in Halmos's Hilbert Space Problem Book): take a_n to be the bilateral weighted shift with weights 1 everywhere except at the 0-coordinate, which has weight 1/n. The limit point a is similar except that the weight at the 0-coordinate is 0. Then each a_n has the unit circle for its spectrum, but a is not ... 2 The proof is making (liberal) use of the following identifications. If P is a finite-rank projection in B(H) , say of rank n, then PH\simeq\mathbb C^n and P\, B(H)\,P\simeq M_n(\mathbb C) . To see this last identification, construct a basis \{e_i\} of H such that e_1,\ldots,e_n form a basis of PH. Construct the corresponding matrix units ... 1 This \mu is a probability distribution on [0,\infty) (from the case n=0). Let X be a random variable with that distribution. Then n^n are the moments of X, and the moment generating function of X is$$ \mathbb E[e^{tX}] = \sum_{n=0}^\infty \frac{n^n}{n!}\;t^n = \frac{1}{1+W(-t)} $$where W is the Lambert W function. So our measure is the ... 0 For a general normed space it is enough to use that the norm is weakly sequentially lower semicontinuous, then$$ ||u|| \leq lim\ inf\ ||u_k|| \leq lim\ sup\ ||u_k|| \leq R.$$Also notice (it is not clear from the statement of the problem that this fact is known to you) that every weakly convergent sequence is bounded. To see this you may use the canonical ... 1 The problem is that the closed unit ball B is not compact when X is infinite dimensional. However, using the definition of equi-continuity, we get \delta such that if \lVert x-y\rVert\lt \delta, then for each \alpha, \lVert T(x)-T(y)\rVert\leqslant 1. In particular, with y=0, we get \lVert T_\alpha(x)\rVert\leqslant 1 for each \alpha if ... 0 To directly show that V maps bounded sets to precompact sets, the only idea I have would be to replicate most of the proof of Ascoli's theorem. We can show it semi-directly, though, by exposing V as the norm-limit of operators with finite-dimensional range. For 1 \leqslant k < n, let$$\lambda_{n,k}(f) = \int_0^{k/n} f(t)\,dt,$$and ... 1 Define F(t) = \int_0^x |f(t)|^p\,dt. This is a continuous increasing function on [0,1]. By the intermediate value theorem, for every i=1,\dots,n-1 there exists x_i\in (0,1) such that$$F(x_i)=\frac{i}{n} F(1)$$These are the points you choose. Indeed,$$\int_{x_{i-1}}^{x_i} |f(t)|^p\,dt = F(x_i)-F(x_{i-1}) = \frac{i}{n} F(1)-\frac{i-1}{n} F(1) = ...

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Since $\|f_n-f\|_p\to 0$, for every $k\in \Bbb N$, there exists $n_k\in \Bbb N$, such that $\|f_{n_k}-f\|_p\le \frac{1}{2^k}$. Let $$g:=|f|+\sum_{k=1}^\infty |f_{n_k}-f|.$$ By definition, $g$ is measurable, and $g\ge |f|+|f_{n_k}-f|\ge|f_{n_k}|$ for every $k\in \Bbb N$ . Moreover, by Minkowski's inequality, $$\|g\|_p\le \|f\|_p+\sum_{k=1}^\infty ... 0 H^{-1}=(H^1_0)^*, so −\Delta:H^1_0(Ω) \rightarrow H^{−1}(Ω) means -\Delta u \in H^{−1}(Ω) for any u \in H^1_0(Ω), and that is obviously right. Because -\Delta u \in H^{−1}(Ω) means (-\Delta u,v) is a scalar, for any v \in H^1_0(Ω). this also satisfies the definition of weak solution. Therefore, \Delta u \in L^2(\Omega), you mean this ... 1 If the domain of our functions is \mathbb{R}, you can show in fact that there are elements of (L^\infty)^\ast which are not in L^1, but given every g\in L^1 we can associate it with a linear functional \psi_g:L^\infty\to \mathbb{R} by \psi_g(f)=\int fgdx, so in fact we have L^1\subset (L^\infty)^\ast, and the containment is strict. To see that ... 1 Since there is no information about how \nabla u changes with respect to t, we can't hope to obtain any cancellation in |\nabla u(t+h,x) - \nabla u(t,x)|. Just estimate it by |\nabla u(t+h,x)| + |\nabla u(t,x)|. Using the notation U(t) = \|\nabla u(t,\cdot)\|_{L^2} and the Cauchy-Schwarz inequality, we estimate the given integral by ... 0 Yes, it is possible to do this, but you have to think careful about the space of functions that you are working with, and what the topology on that space is. It is often possible to solve certain PDE by considering their solutions to be fixed points of a continuous map on a convex subset of a Banach space, such as a Sobolev space. Existence would follow from ... 0 IF T-S is bounded on Y, and if T^{-1} : X\rightarrow Y is bounded, then (T-S)T^{-1}=I-ST^{-1} is bounded on X, and \|I-ST^{-1}\| < 1 by your assumptions. Therefore, ST^{-1}=I-(I-ST^{-1}) is invertible in \mathcal{L}(X). And (ST^{-1})^{-1}ST^{-1}=ST^{-1}(ST^{-1})^{-1}=I. It follows that S is surjective from the second equality. And, ... 0 From the inequality \lVert T-S\rVert \lVert T^{-1}\rVert < 1, it follows that$$R = (I - T^{-1}(T-S))$$is an invertible bounded operator. Since TR = T(I - T^{-1}(T-S)) = T - (T-S) = S, we get the representation S^{-1} = R^{-1}T^{-1} in the bounded case. Now we can verify that R^{-1}T^{-1} is also the inverse of S when S is unbounded. Since ... 0 Yes. To be explicit, let T:W\to C be a bounded linear operator (it need not be injective). For every linear functional f\in C^* we have f\circ T\in W^*. Since u_n converge weakly, f(T(u_n))\to f(T(u)). But this says precisely that T(u_n) converge to T(u) weakly. 1 Yes. If the embedding j \colon W \hookrightarrow C is continuous when the two spaces are endowed with their respective norm topologies, it is also continuous when both spaces are endowed with their weak topologies, and that means weakly convergent nets are mapped to weakly convergent nets. In particular, if the weakly convergent net is a sequence, its ... 2 I'm not entirely sure if this is what you wanted but here is a sketch. Let me know of any confusions: We have V a finite dimensional vector space, say dimension n, and so the subspace W is also finite dimensional, say of dimension n-k. Let's choose a basis for W, say (w_1, \ldots, w_{n-k}), and then extend this to a basis of V, say (w_1, ... 0 The essential numerical range contains the essential spectrum and the later is never empty. 0 Hint: Show that X=L. Start by picking a point x in X with \|x\|=1, then consider the point (N-1/2)x \in NB\backslash (N-1)B. What can you conclude about the unit ball of X? What does this imply? 1 Vectors in W^\ast are just linear maps W \to \mathbb F and f^\ast takes a map W \to \mathbb F to the composition V \to W \to \mathbb F. So the second condition states that there exists a linear map l\colon W \to \mathbb F such that the composition l\circ f is zero but l(w) = 1. One way to do this question is to make a clever choice of ... 1 You can derive such an inequality from the Gagliardo-Niremberg inequality. 0 You can not have an inequality of the form$$ |u|_{H^1}^2\le C(|u|_{H^2}+\|u\|_{L^2}) $$Assume that this is true and replace u by Mu, where M>0 constant. Then the left hand side is {\mathcal O}(M^2) while the right-hand side is {\mathcal O}(M). In your argument, try instead M\sin kx, for k,M large. 4 Use integration by parts as follows$$\langle f, e_n \rangle=\frac{1}{\sqrt{2\pi}}\int\limits_{-\pi}^{\pi} x e^{inx}\, dx \\= \frac{1}{\sqrt{2\pi}}\left[\left.x\cdot \frac{e^{inx}}{in}\right|_{-\pi}^{\pi}-\int\limits_{-\pi}^{\pi} \frac{e^{inx}}{in}\cdot 1\, dx\right]$$Added The integral will be$$c_n=\frac{1}{\sqrt{2\pi}}\cdot \frac{2\pi\cos n\pi}{in} \\ ...

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I think the answer is negative. An example would be the sequence $\mu _n = \delta _{\frac12 - \frac1n}-\delta _{\frac12 + \frac1n}$, which converges in weak* topology to the trivial measure $\mu$, $\mu ([0;1]) = 0$.

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The name functional was originally used to refer to functions whose arguments were functions. In this sense an operator on a vector space of functions would be a functional. Looking around it appears that the first use of the word functional was in Jacques Hadamard's book "Leçons sur le calcul des variations" which translates to "Lessons in the calculus of ...

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When dealing with infinite dimension we have to be careful with our notation since although the sets do not coincide, the closures (with respecto to any of the topologies below) do coincide. Denoting with $\otimes _a$ the algebraic tensor product and without the subindex the closure w.r.t. any of the weak, $\sigma$-weak, strong, $\sigma$-strong, strong*, ...

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Let $K$ denote the scalar field. Consider $F\colon X \to K^n$ given by $$F(x) = \begin{pmatrix}L_1(x)\\ L_2(x)\\ \vdots \\ L_n(x)\end{pmatrix}.$$ Let $R = \operatorname{im} F \subset K^n$. We have an induced isomorphism $$\tilde{F}\colon X/\ker F \xrightarrow{\sim} R.$$ Since $\bigcap\limits_{k=1}^n \ker L_k = \ker F \subset \ker L$, we have an induced ...

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I have a proof in the case where $X$ is reflexive. Suppose that $L$ is not a linear combination of the $L_i$'s. Let $C = \{\sum_{i=1}^n t_i L_i : t_i \in \mathbb{R}\} \subseteq X^*$. Then $C$ is a closed convex subset and $C \cap \{L\} = \emptyset$ by assumption. So by geometric Hahn-Banach, there exists $\xi \in X^{**}$ such that $\xi(C) \subseteq ... 0 Let$E=C([0,1],\Bbb R)$(continuous maps frome$[0,1]$to$\Bbb R$) with$\|.\|_{\infty}$and$\|.\|_1$and identical mape :$f \mapsto f$frome$(E,\|.\|_{\infty})$to$(E,\|.\|_{1})$0 You can see that$\forall x \in[0,1],0\le \gamma(x) \le 1$,$\gamma(0)=1$and$\gamma(1)=1$. From that, you can deduce that$\forall n\in\mathbb N,\forall x \in[0,1], 0\le\Psi_{n}(x)\le 1$,$\Psi_{n}(0)=1$and$\Psi_{n}(1)=1$. So assuming$\Sigma$is well defined on$]0,1[$,$\Sigma(0)=1$,$\Sigma(1)=1$and$\forall x \in[0,1],0\le \Sigma(x) \le 1$, ... 0 In finite dimensions, all norms are equivalent. (For example, see Any two norms on finite dimensional space are equivalent.) This means the following: if$\Vert \cdot \Vert_1$and$\Vert \cdot \Vert_2$are the two norms on the space$X$, there are constants$a>0$and$b>0$such that $$a \Vert x \Vert_1 \leq \Vert x \Vert_2 \leq b \Vert x \Vert_1,$$ ... 1 Every finite-dimensional normed space of dimension$n$over$\mathbb{F}$is isomorphic to$\mathbb{F}^n$, and hence complete (assuming$\mathbb{F}=\mathbb{C}$or$\mathbb{R}$). Your student's argument works! The space that this norm defines (in$n$dimensions) is somtimes called$l_1^n$(which is complete). You can also apply the same norm (infinite sum) to ... 1 The proof is correct. The first embedding$W^{1,\infty}(I) \hookrightarrow C^{0,1}(I)$is actually an isomorphism; depending on how you norm$W^{1,\infty}(I)\$, it may be an isometric isomorphism. I would not call this (easy) fact an application of Morrey's lemma myself, but on the other hand it makes sense to classify it as an endpoint case of Morrey's ...

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