# Tag Info

0

Convolution really comes out of gathering like powers of things such as powers of exponentials or of powers of a complex variable. \begin{align} \sum_{n=-\infty}^{\infty}a_n e^{in\theta}\sum_{n=-\infty}^{\infty}b_ne^{in\theta}&=\sum_{n=-\infty}^{\infty}\left(\sum_{j+k=n}a_j b_k\right)e^{in\theta} \\ & = ...

2

Some important tricks/theorems: The spectrum is non-empty for Banach algebras (over $\mathbb{C}$) The spectral radius formula $$r(a) = \lim \|a^n\|^{1/n}$$ tells you that if $a$ is nilpotent, then $\sigma(a) = \{0\}$ If $A$ is commutative, then $$\sigma(a) = \{\tau(a) : \tau \in \Omega(A)\}$$ where $\Omega(A)$ denotes the set of non-zero multiplicative ...

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No, this is false. Let $f\in H^\infty(\mathbb{D})$ be a function that is continuous on $\partial D\setminus \{1\}$ and does not have a limit as at $z\to 1$. (E.g., $f$ could be a conformal map onto a domain one with one nontrivial prime end.) If $f=g+\phi(z^2)$ with $g\in A(\mathbb{D})$, then $\phi$ must be discontinuous at $1$; but then $\phi(z^2)$ is also ...

3

That's a good proof. Here is another one. Assume that $a^m=0$. If you know that $$\sigma(a)=\{f(a):\ f \text{ is a multiplicative functional }\},$$ then for any such $f$ we have $f(a)^m=f(a^m)=f(0)=0$. So $f(a)=0$, and then $\sigma(a)=\{0\}$. Yet another proof, without machinery: if $a^m=0$, then $1-a$ is invertible: indeed, the inverse is ...

1

Yes, that is one way to look at it. Here's another: even before the Gelfand transform, we already know from Lemma 1.2.4 that $\sigma(a)$ is compact. Here it is only proven for unital Banach algebras, but that is simply because Murphy only defines the spectrum of an element in a non-unital algebra on page 13, at the very end of section 1.2. (Think about it: ...

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No, not in general. For instance, suppose $A=\mathbb{C}$ with the standard norm. Then (via a change of basis $(a,z)\mapsto (a+z,z)$) we can identify $A_+$ with $\mathbb{C}^2$ with coordinatewise multiplication, and your norm with $\|(a,b)\|=(|a-b|^p+|b|^p)^{1/p}$. Now consider the elements $x=(1,0)$ and $y=(1,1/2)$. We have $\|x\|=1$ and ...

2

Well, nothing special: write $C \in M_{mn}(\mathbb C)$. Let $C_{ab} \in M_{mn} (\mathbb C)$, where $a,b \in \{1, 2, \cdots, n\}$ so that $$(C_{ab})_{ij} = \begin{cases} C_{ij} & \text{if } (a-1)m +1 \le i\le am, (b-1)m+1\le j\le bm,\\ 0 & \text{otherwise.}\end{cases}$$ Abusing notations, we also consider $C_{ab} \in M_m(\mathbb C)$. Then C = ...

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.

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Let $f : V \to W$ be a function between two Banach spaces. Then by definitnion, the Frechet derivative at $x$ is the only bounded (=continuous) linear operator such that... Hence $\nabla f(x) \in L(V,W)$. So $\nabla f(x) \in V'$ if and only if $W = \Bbb R$ (or $\Bbb C$). For your question 2), if $f:V\to W$, then $\nabla f(x) : V\to W$, so it doesn't ...

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Note that $a-\lambda$ commutes with $be^{i\lambda}$. In any ring, if $x$ and $y$ commute and the product is invertible, then each is invertible. Proof: suppose that $zxy=xyz=I$. Then $zyx=zxy-I$, so $x$ has a left inverse. From $xyz=I$ we know that $x$ has a right inverse. Then $x$ is invertible.

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