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## Hot answers tagged functional-analysis

9

Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis? It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2. I tried to solve the exercise, but I'm not quite sure that I understood it. Let ...

7

Assume without loss of generality that $a_1\ne0$ and consider, for every $n\geqslant1$, $$b_n=\frac{a_n}{A_n},\qquad A_n=\sum_{k=1}^na_k^2.$$ Then $$\sum_na_nb_n=\sum_n\frac{a_n^2}{A_n},$$ which diverges by a standard result. On the other hand, $A_n^2\geqslant A_nA_{n-1}$ and $a_n^2=A_n-A_{n-1}$ hence, for every $n\geqslant2$, $$... 6 The answer is: Not in general. For example, \Omega=(0,2\pi), u_n(x)=\sin nx. Then u_n\rightharpoonup 0, since$$ \int_0^{2\pi} f(x)\,\sin nx\,dx\to 0=u, $$for all f\in L^2[0,2\pi]. Meanwhile$$ v_n(x)=u_n^2(x)=\sin^2 nx=\frac{1}{2}-\frac{\cos (2nx)}{2}\rightharpoonup \frac{1}{2}=v, $$and u^2\ne v. Note. If \{u_n\} is bounded and ... 5 No, the argument is not valid. Let me give two examples: Let X be a metric space. Let Y\subseteq X be a subset. Then if (y_n) is a Cauchy sequence in Y such that y_n \to x_0 \in X\setminus Y, then Y is not complete. This is because for every y\in Y, 0 < d(y,x) \leq d(y,y_n) + d(y_n,x) by the triangle inequality. So (y_n) cannot ... 5 The simplest counterexample is the non-zero constant function$$ f(x)=1. $$If g\in C_0^1(\mathbb R), then \lim_{|x|\to\infty}g(x)=0, and hence$$ \lim_{x\to\infty}|f(x)-g(x)|=\lim_{x\to\infty}|1-g(x)|=1. $$Thus$$ \|f-g\|_\infty=\sup_{x\in\mathbb R} |f(x)-g(x)|=1, $$and therefore f can not be approximated by C_0^1 functions. 5 Since you are not assuming integrability of mixed partials, this is not so much a Sobolev embedding situation, but more of elliptic regularity result. Luckily everything is in L^2, which leads to conclusion quickly. Consider the Fourier transform \hat f(\eta,\xi) where (\eta,\xi)\in\mathbb R^2. Your assumptions imply$$|\eta|^k \hat f \in L^2, \quad ...

5

(1): The magical ingredient that makes Hilbert spaces behave so much nicer than general Banach spaces is the parallelogram identity: Lemma: Let $H$ be a Hilbert space and $x,y\in H$. Then $$\|x+y\|^2+\|x-y\|^2=2(\|x\|^2+\|y\|^2)$$ Proof: Verify using $\|x\|^2=\langle x,x\rangle$ and straightforward calculation. $\square$ Theorem: Let $H$ be a ...

5

For any $f \in L^1$ we have $$\lim_{\omega \to \infty} \underbrace{\int_{\mathbb{R}} f(x) \cdot e^{-\imath \, x \omega} \, dx}_{=:\hat{f}(\omega)} = 0, \tag{1}$$ this result is known as Riemann-Lebesgue-Lemma. Since $f \in L^1$ we can choose a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions such that $f_n \stackrel{L^1}{\to} f$. Since ...

4

Yes, we could also define the strong resp. weak operator topologies as the initial topology with respect to the evaluation maps $p_x$, where $Y$ is endowed with the strong resp. weak topology. For the weak operator topology that is obvious from the transitivity of initial topologies, since the weak topology is just the initial topology with respect to the ...

4

Let $F:X\to\mathbb R^n$ defined as follows $$F(x)=\big(f_1(x),\ldots,f_n(x)\big).$$ Clearly $F$ is a linear transformation and its range $Y$ is linear subspace of $\mathbb R^n$. We shall show that the range is the whole $\mathbb R^n$. If it is not, then we can find a vector $v\in\mathbb R^n$, such that $v=(v_1,\ldots,v_n)\perp Y$, i.e. $$v\cdot y=0, ... 4 I do not think that \mathscr D(\Omega) is sequential. On the other hand this is probably not used: By definitionn of the locally convex inductive limit topology of \mathscr D(\Omega)= \lim X_n (where X_n are the Frechet spaces of smooth functions with support in K_n for a compact exhaustion) a linear map with values in any locally convex space is ... 4 The functions belonging to L^p are in general not continuous. And, to be precise, the elements of L^p are not even functions, but equivalence classes of functions. First we start with the space \mathscr{M} of measurable functions. Measurable functions can be pretty wild, but they have enough regularity that one can handle them. For a measurable ... 4 Bessaga showed something stronger, but only for Hilbert spaces. Generalization to certain Banach spaces (i.e., those which are linearly injectable into some c_0(\Gamma)) was given by Dobrowolski: In particular, in 1966 C. Bessaga [1] proved that every infinite-dimensional Hilbert space H is C^\infty diffeomorphic to its unit sphere. The key to prove ... 4 Let F\subset T(X) be a closed (in Y) subspace. Then E = T^{-1}(F) is a closed subspace of X, and T\lvert_E \colon E \to F is a compact surjective operator. Since F is closed in Y, the open mapping theorem implies that F is finite-dimensional. So: the image of a compact operator cannot contain an infinite-dimensional Banach space. 4 Under the assumption that \epsilon is constant across a, then no additional assumptions are needed. Suppose that x_a is the fixed point of f_a, and choose e>0. Then there is a \delta such that for all b within \delta of b, you have e\epsilon\geq \vert f_b(x_a)-f_a(x_a)\vert=\vert f_b(x_a)-x_a\vert. Now, repeatedly apply f_b to ... 4 The main problem is that, while the L^1-Norm is in fact a norm on C^0[0,1], it is only a seminorm when considering measureable functions. This is why one usually to considers equivalence classes of summable functions. If the initial example doesn't compel you (as the nonunique limit function is a.e. zero and is in the same L^1-equivalence class as ... 3 In Banach space X a sequence \{f_n\} converges weakly to f if$$ \varphi(f_n)\to\varphi(f), $$for all \varphi\in X^*, where X^* is the dual of X. In the case of Hilbert space H, every element of the dual space is realized by an element of H (Riesz Representation Theorem). Thus f_n\to f weakly if and only if$$ \langle ...

3

Let me restate your question to make sure I understand. You consider the space $C^0[0,1]$ of continuous functions on $[0,1]$ equipped with the $L^1$ norm: $$||f|| = \int_0^1 |f(x)|\,dx$$ You are asking: is the sequence $f_n(x) = x^n$ a Cauchy sequence in this space without a limit? The answer is no. The sequence is a Cauchy sequence because: $$||f_n - ... 3 A condition is "T is dense in [0,1] for the usual topology". Indeed, in this case, x(t)=0 for each t\in T implies by continuity the same for t in the closure of T. If T is not dense in [0,1], then [0,1]\setminus \overline T contains a non-empty open set, hence a small interval. We then construct a continuous function which doesn't vanish ... 3 A solution, by my professor: Consider the closed unit ball in \ell^{\infty}(I). The extreme points of this are functions f: I \to \{-1, 1\}. Notice that if \phi: \ell^{\infty}(I) \to \ell^{\infty}(J) is an isometric isomorphism, then \phi takes extreme points to extreme points. We may assume without loss of generality that \phi(\chi_I) = \chi_J, ... 3 Let H=L^{2}[0,\pi]. The subspace D=\mathcal{C}^{\infty}_{c}(0,2\pi) consisting of infinitely-differentiable functions on (0,2\pi) which are compactly supported in (0,2\pi) is a dense subspace of H. Let \mathcal{D}(A) be the domain of twice absolutely continuous functions on [0,2\pi] for which$$ f'\in H, f'' \in H, f(0)=f(\pi)=0. ...

3

Liouville's theorem (the one about bounded entire functions) is in complex analysis only proved for complex-valued functions. The generalisation to $\mathbb{C}^n$-valued functions is immediate, but for general Banach-space-valued (or Banach-algebra-valued) functions, it needs to be proved before it is used. The proof in that general case is by reducing it to ...

3

I think your prof is right, and this has nothing to do with series and/or norms. The space of polynomials of degree less than a fixed number is finite-dimensional and in such space the operator $D$ is nilpotent and satisfies your prof's formula. As another way to see that your trick works only for polynomials, note that the solution to your equation is ...

3

You ask what is the definition of a positive element in $E$; the element $a \in E$ is positive if $a$ is positive when we think of it as an element in $A$. But an operator system is more than that. It also allows us to define an order structure on $M_n(E)$ for each $n$ and to say whether each matrix $[a_{ij}]\in M_n(E)$ is positive or not. The statement "an ...

3

$\Omega$ is bounded, so there is a K such that $\Omega \subset \mathbb{R}^n\times (-\infty,K)$. Then it's basically John's hint from the other question: For $u \in C_0^\infty(\Omega)$, we have $$u(x,0) = - \int_0^K \frac{\partial u}{\partial x_{n+1}}(x,t)\,dt.$$ Hölder's inequality gives $$\lvert u(x,0)\rvert \leqslant \left(\int_0^K \left\lvert ... 3 For \lVert\xi\rVert_X = 1, on the line through 0 and \xi, the minimal value of$$h_\xi(t) = \frac{\lVert t\xi\rVert_X^2}{2} - L(t\xi) = \frac{1}{2}t^2 - t\cdot L(\xi)$$is -\frac{1}{2}L(\xi)^2, attained at t = L(\xi). Thus if$$\frac{\lVert x_0\rVert_X^2}{2} - L(x_0) \leqslant \frac{\lVert x\rVert_X^2}{2} - L(x)\tag{1} for all $x\in X$, then ...

3

A topological space is intuitively a space with a notion of "nearness". These spaces are very general, and some of them are called Hausdorff spaces because they satisfy a special property. A metric space is a space with a notion of distance, and this distance will always imply a notion of nearness: In other words, every metric space is a topological space. ...

3

No, it's not. You're confusing the notion of convergence of a sequence of numbers with convergence of a sequence of elements of $\ell^\infty$, that is to say, a sequence of sequences. Of course, for any $(x_n)_n\in c_0$ we have $x_n\to 0$, but this is not the same zero as $0\in c_0$, as the former is a number, and the latter is a sequence, comparison ...

3

There is a simple but interesting result which says "every two norms on a finite dimensional vector space are equivalent" and then you can conclude the reasoning proving that equivalent norms induce equivalent topologies on the same space. look here for further details

3

Another proof to see it is not metrizable: $X$ is separable: every set $[a,b)$ with $a < b$ contains a rational (already in $(a,b)$), so $\mathbb{Q}$ is dense and countable. But $X$ is not second countable: let $\{ B_i: i \in I\}$ be a base for $X$. For each $x$ in $X$, we pick $i(x) \in I$ such that $x \in B_{i(x)} \subset [x, x+1)$, by the definition of ...

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