Functional calculus allows the evaluation of a function applied to a linear operator or a matrix. The function could be a polynomial, a holomorphic function, a continuous function or a measurable function defined on the spectrum of an operator or a Banach algebra. Functional calculus is a basic and powerful tool in the spectral theory of operators and operator algebras and is part of functional analysis.
Questions tagged [functional-calculus]
368 questions
2
votes
1 answer
Derivative of fractional iteration.
On wikipedia, I know it's not the best reference but still, there is an article about fractional iterations. I attach the fragment I want to ask about.
How do we calculate derivative of fractional iteration? I know this formula works for natural…
1
vote
0 answers
Riesz Functional Calculus for Continuous Functions
Let $A$ be a bounded normal operator in a Hilbert space. Then we know there exists a continuous functional calculus defining $f(A)$ for $f\in C(\sigma(A))$ in a reasonable way. But there is also the Riesz functional calculus for functions $f$ that…
Friedrich Philipp
- 4,051
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functional calculus and laplacian
hi i have a question:the operator $-\Delta$ with domain $H^{2}(R))$ the question is:
if $f\in C_{0}^{\infty}(R)$ prove that f(T) is well defined?
clearly the answer is by using functional calculus but as i know if f \ in C(spectrum($-\Delta$)) the…
RIM
- 51
0
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1 answer
Spectrum of isometry not in circle
I have a reasoning that is wrong but I don't understand why.
Suppose we have a unital $C^*$-algebra $A$ and $v \in A$ is a proper isometry, so $v^*v = 1$ but $v v^* \neq 1$.
Take the function $f: \mathbb{C} \to \mathbb{C}: y \mapsto…
Emiel Lanckriet
- 189
0
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1 answer
Automorphism of the $C*$ algebra $C(\mathbb{T})$
Let $C(\mathbb{T})$ denotes the C* algebra of continuous functions on the unit circle. Let $\alpha: C(\mathbb{T}) \to C(\mathbb{T})$ be defined by $\alpha(f)(z)=f(e^{-2\pi i\theta} z)$. I need to prove that $\alpha$ is an automorphism. It is clear…
budi
- 1,780
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Find min-max and concavity when first derivative is a Transcendental Function
I'm trying to sketch the function $$f(x)=\frac{\ln(5x^2+x)}{2-3x}$$ .
When I do the first derivative I get $$\frac{d}{dx}\bigg(\frac{\ln(5 x^2 + x)}{2 - 3 x}\bigg) = \frac{10 x + 1}{(2 - 3 x) (5 x^2 + x)}+\frac{3\ln(5 x^2 + x)}{(2 - 3 x)^2}$$…
Stramike
- 11
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Derivative of a functional
If a functional is defined as follows $$\mathcal{F}[\rho(x,t)]=\int_0^{t}w(x,\tau)\rho(x,t)d\tau$$ what is the derivative of this functional with respect to $x$ i.e., $$\frac{\partial}{\partial x}\Big[\mathcal{F}[\rho(x,t)]\Big]$$
Rhinocerotidae
- 205
0
votes
1 answer
Functional derivatives of metric tensor
What is wrong in the following?
\begin{equation*}
g_{\mu\nu}(x)g^{\mu\nu}(x)=D \\
\frac{\delta}{\delta g_{\alpha\beta}(x)}D=0 \\
\frac{\delta}{\delta g_{\alpha\beta}(x)}(g_{\mu\nu}g^{\mu\nu})=2g^{\alpha\beta}(x)=0
\end{equation*}
The above…
Giuseppe
- 249