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## Hot answers tagged functional-analysis

6

Yes, this can be done without making the $C^*$-algebra concrete in $B(H)$. All we need is that $z=0$ iff $z^*z=0$, which follows from the $C^*$-identity $\|z^*z\|=\|z\|^2$. Lemma Let $a\in A$. The following assertions are equivalent. 1 - $p=a^*a$ is idempotent, i.e. $p^2=p$. 2 - $aa^*a=a$ 3 - $a^*aa^*=a^*$ 4 - $q=aa^*$ is idempotent. Proof ...

6

W.l.o.g. $\alpha \ge \beta$. $$\alpha\|x+y\| = \|(\alpha-\beta)y + (\alpha x + \beta y) \| \le (\alpha-\beta)\|y\| + \|\alpha x + \beta y\| \le (\alpha-\beta)\|y\| + \alpha\|x\| + \beta \|y\| = \alpha(\|x\|+\|y\|)$$ Since the first-quantity = last-quantity, all the inequalities must be equalities. In particular: $$(\alpha-\beta)\|y\| + \|\alpha x + \beta ... 5 Well, you could have done the shortcut: if f is continuous over [a,b], then so is |f|^p, so it is integrable over the compact interval [a,b]. However, it is not closed. Consider for simplicity [a,b]=[-1,1] and let f be the signum function on it: f(x)=\displaystyle\frac x{|x|} if x\ne 0 and f(0)=0. This is not continuous, but can be ... 4 Here is the refrence to the original article, see theorem 83. The main idea of the proof is the following. Let Z be a compact Hausdorff space. For a given p\in Z denote M_Z(p)=\{f\in C(Z):|f(p)|=\Vert f\Vert\}. Also denote \mathcal{M}_Z=\{M_Z(p):p\in Z\}. Given M\in\mathcal{M}_Z we can always recover its point p\in Z. Once we have a ... 4 Suppose we had a u satisfying u \ast f = f for all f \in L^1(\Bbb{R}^d). Then in particular u \ast u = u . Then taking Fourier transforms we get \hat{u}(x)\hat{u}(x) = \hat{u}(x) for all x \in \Bbb{R}^d. Now the Fourier transform of an L^1 function is always continuous and bounded. So for every x, either \hat{u}(x) = 0 or else is not-zero. ... 4 Since P+Q=(P+Q)^2, we get P+Q=P+Q+PQ+QP, so PQ+QP=0. Multiplying by P both left and right, we get 2PQP=0, so PQP=0. Now we use that P^*=P, Q^*=Q:$$ 0=PQP=PQQP=P^*Q^*QP=(QP)^*QP, $$so QP=0. Then 0=\langle QPx,y\rangle=\langle Px,Qy\rangle for any x,y, showing that the ranges are orthogonal. 3 For x,y \in H, consider the compact operator$$ \Theta_{x,y}(z) := \langle z,x\rangle y $$If A\Theta_{x,y} = \Theta_{x,y}A for all x,y\in H, we have$$ \langle z,x\rangle A(y) = \langle A(z),x\rangle y \quad\forall x,y,z\in H $$Taking z = x non-zero, we have A(y) = \alpha y for all y\in H where$$ \alpha = \frac{\langle A(x),x\rangle}{\langle ...

3

Another example, with functions $f_n$ in $C^\infty([0,+\infty))$: $$f_n(x)=\mathrm e^{-nx}.$$ Then $\|f_n\|_\infty=1$ and $\|f_n\|_p=(np)^{-1/p}\to0$ hence the inequality $\|f\|_\infty\leqslant C\|f\|_p$ for every $f$ in $C^\infty([0,+\infty))$ and for some finite constant $C$ independent of $f$, is impossible.

3

If you want to understand well topological vector spaces (TVS), and learn all the basic theorems of functional analysis in their most abstract version, which is usually in the context of TVS and not of normed spaces, I am afraid there is no speedy way. (No royal way to geometry, as the ancient greeks said.) I would recommend the book of W. Rudin, Functional ...

3

I have a very similar reference: The German "Funktionalanalysis" (functional analysis) by D. Werner: He defines for a (complex) locally convex topological vector space $X$ with dual $X'$: $$A^\circ = \{ x' \in X' : \Re \langle x' , x\rangle \le 1 \; \quad\forall x \in A\}.$$ (In the real case, you can simply drop the "$\Re$".) He calls this set "Polare", I ...

3

The author wrote One can check that the distance is $\frac12$. That doesn't (necessarily) mean it's obvious. Let's see what we can find. For $a = 0$, we obviously have $\lVert e_1 - a\rVert_1 = 1$, so let's try to find something closer. Write $$a = c_1\cdot e_1 - \sum_{k=2}^\infty c_k\cdot e_k,$$ where not all $c_k = 0$. Then to have $a\in A$, we ...

3

For this to hold it is crucial that $H$ is a complex Hilbert space since on $\mathbb{R}^2$ a rotation $T$ by $\pi/2$ has no eigenvalues, yet $T^2 = -I$ is compact and self-adjoint. The spectral theorem for compact self-adjoint operators yields an eigenvector $w$ of $A^2$ with real eigenvalue $\lambda$. Without loss of generality we can rescale $A$ ...

3

Since $A^2$ is compact, it has an eigenvalue $\alpha$. As $A^2$ is self-adjoint, $\alpha$ is real. Let $\lambda\in\mathbb C$ with $\lambda^2=\alpha$. As $\alpha$ is an eigenvalue for $A^2$, we have that there exists nonzero $v$ with $(A^2-\alpha I)v=0$. Then $$0=(A^2-\alpha I)v=(A-\lambda I)(A+\lambda I)v=0.$$ If $(A+\lambda I)v=0$, then $-\lambda$ is ...

3

To answer your questions, we use the fact that an operator on a Hilbert space is compact if and only if it maps some orthonormal basis to a sequence converging to $0$. Question 1: No. Consider the operator on $\ell^2$ given by $$T(\{x_k\}) = \{k^{-1}x_k\}$$ and denote by $e_j$ the standard basis. $T$ is obviously self-adjoint an $Te_j = j^{-1}e_j \to 0$ so ...

3

You can "almost explicitly" give two different extensions. Let $E \subset \ell^\infty$ be the subspace of sequences $x\in\ell^\infty$ such that $e_0(x) := \lim\limits_{n\to \infty} x_{2n}$ exists, and $O \subset \ell^\infty$ the subspace of sequences $x\in\ell^\infty$ such that $o_0(x) := \lim\limits_{n\to\infty} x_{2n+1}$ exists. Evidently $c \subset E\cap ... 2 Hint: the following statement solves your problem: Statement: If$P_1,\dots,P_n\in\mathcal H$are orthogonal projections such that$\sum_{i=1}^n P_i$is again an orthogonal projection, then$P_i$are pairwise orthogonal, i.e.$P_iP_j=0,\ i\neq j.$Edit. Proof. If$Q,R$are projections such that$Q\geq R$then$Q\supseteq R.$Indeed, let$x\in Ran R.$... 2 The meaning of the first hint is: if$x_n$happens to be in the span of some previously chosen vectors, a small perturbation will get it out of there. This is possible precisely because a proper subspace has no interior points. An interior point wouldn't be able to get out of the set with an arbitrarily small perturbation. The reason you did not use the ... 2 In general you would use the holomorphic functional calculus $$f(T)=\frac1{2\pi i}\,\int_\Gamma \,f(z)\,(z-T)^{-1}\,dz,$$ where$\Gamma$is a curve in the complex plane with the spectrum of$T$inside the region it delimits, and$f$is an analytic function on that region. Now, if$H$is finite-dimensional, things are way easier. Then you have ... 2 It is perfectly valid to arrive at equation$(4)$the way you have. Doing so, however, presents difficulties in applying the Fundamental Lemma of Calculus of Variations. The Fundamental Lemma says that for some function$f(x)$which is$k$times differentiable on some interval$[a, b]$, if we have $$\int_{a}^{b} f(x)h(x)\,dx = 0$$ ... 2 For general$A$, it is not true even in finite dimension in the$1\times1$case: because that would imply the inequality$|\log(1+t)|\leq C|t|$, and we can take$t=-1+\varepsilon$to get the (false) inequality$|\log\varepsilon|\leq C|1-\varepsilon|$for all$\varepsilon>0$. For$A$positive, though, if the eigenvalue sequence of$A$is$\{\lambda_j\}$, ... 2 I think the averaging technique using the Haar measure is crucial to such an argument. For the proof using Haar measure, proceed as follows: Let$\Vert . \Vert$be the norm on$X$. Define the new norm $$\Vert v \Vert_{G}=\frac{1}{\vert G \vert} \int_{G}\Vert \pi(g)v \Vert\ d\mu$$ One can easily verify that this is a norm and is equivalent to$\Vert . ...

2

Let $\mathbb K$ be either $\mathbb R$ or $\mathbb C$. If your spaces are not normed, then you don't need Hahn-Banach. Just take any $x\in E\setminus F$. Now consider the subspace $F'=F+\mathbb K x$. It is a direct sum, as $F\cap\mathbb K=0$. Define a linear functional $f:F'\to\mathbb K$ by $$f(z+\lambda x)=\lambda,\ \ \ z\in F,\ \lambda\in\mathbb K.$$ ...

2

A Taylor expansion would be kind of circular, since a Taylor expansion requires differentiability, and that is what you want to prove. However, noting that $f(u) = \lVert u\rVert_H^2 = \langle u,u\rangle_H$, it is not difficult to get an expansion of $f(u+h)$ using the properties of the inner product (the one you need is bilinearity). This expansion then ...

2

Take $f(x)=|x-a|^{-1/2p}$. Clearly $f\in L^p[a,b]$, but $f\not\in L^\infty[a,b]$. Then take $$f_\delta(x)=\left\{ \begin{array}{lll} f(a+\delta) & \text{if} & x\in [a,a+\delta), \\ f(x) & \text{if} & x\in [a+\delta,b]. \end{array} \right.$$ Clearly, $f\delta$ belongs to both $L^p[a,b]$ and $L^\infty[a,b]$, and $$... 2 Note that J is the composition of two maps$$ \varphi: \ell^2 \to \ell^1 \text{ given by} (x_n) \mapsto (x_n^2) $$and$$ L: \ell^1 \to \mathbb{C} \text{ given by } (x_n) \mapsto \sum n^{1/n} x_n $$Now check that \varphi is continuous, and$$ \lim_{n\to \infty} n^{1/n} = 1 $$and so \exists M > 0 such that n^{1/n} \leq M for all n\in ... 2 Suppose it weren't so. Then you'd have a subsequence (x_{n_k}) with d(f(x_{n_k}), f(x)) \geqslant \varepsilon for some \varepsilon > 0. By your argument, f(x_{n_k}) has a convergent subsequence f(x_{n_{k_m}}). Since the graph is closed, as you said, we must have f(x_{n_{k_m}}) \to f(x). But that contradicts the assumption that d(f(x_{n_k}), ... 2 The Dirac delta function is not an object in L^2[0,1]. You have, for any k\in\mathbb N\cup\{0\},$$ \int_0^1g_n(t)\,t^k\,dm(t)=\frac1{k+1}\,\frac1{n^{2k+1}}\to0. $$This shows that \int_0^1g_n\,p\to0 for any polynomial p. Now, polynomials are dense in L^2[0,1], so a trivial application of Hölder lets you show that$$ \int_0^1g_n\,f\to0  for all ...

2

Regarding the first question, yes, it is the algebra of polynomials, if you consider polynomials in $z$ and $\overline{z}$. This is also the algebra of trigonometric polynomials, generated by $\cos nt,\, n \geqslant 0$ and $\sin nt,\, n\geqslant 1$. By Weierstraß' theorem, this algebra is dense in $C(S^1)$.

2

The elements of $\ell_\infty$ are infinite sequences. To show $\|x+y\|_\infty\le \| x\|_\infty+\| y\|_\infty$, use the following strategy: a) for every $n$, $|x_n|\le \| x\|_\infty$ and $|y_n|\le \| y\|_\infty$. b) Use the above to estimate $|x_n+y_n|$. c) Now you have an upper bound for $|x_n+y_n|$; the supremum is less than or equal to an upper ...

1

$C_1$ has no interior points means that $X\setminus C_1$ is nonempty, more, every nonempty open set intersects $X\setminus C_1$. From the non-emptyness, we obtain an $x_1 \in X\setminus C_1$, and since $C_1$ is closed, we have a $\rho_1 > 0$ with $B_{\rho_1}(x_1) \subset X\setminus C_1$. For every $0 < \varepsilon_1 < \rho_1$, we know ...

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