## functional analysis – contraction in five dimensional Euclidean space

$$T(u,v,w,x,y)=(v,0.5w,x,y,u)$$

$$S(u,v,w,x,y)=(y,u,v,0.5w,x)$$

Could any one tell me whether S and T are contraction map on any matrix norm/induced vector norm? Are their composition both way i.e ST and TS, contraction? What can we say about Contraction of SSSSTTTT or TTTTSSSS? Thank you for your response.

## Is presenting a message to the user a functional or non-functional requirement?

For my application, I will present the user with a message or quote from a famous person on the completion of a game.

Would this requirement be functional or non-functional?

## functional analysis – Integral operators being trace class

Consider $$T$$, a bounded integral operator on $$L^2(mathbb{R}^2)$$. The integral kernel of $$T$$ is denoted by $$left$$, denoted $$T(vec{x},vec{y})$$ for simplicity (and confusion). I am interested in when such operators are in fact trace class. That is, does there exist any result such as,

$$textbf{(Possible) Lemma.}$$ $$T$$ is trace class (maybe with more assumptions) if and only if
$$int_{vec rin mathbb{R}^2} dvec{r}left(int_{vec xin mathbb{R}^2}dvec{x},|T(vec{r}+vec{x},vec{x})|,right),.$$
is finite.

I have the following result, Proposition 3.3 in this paper https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-159/issue-2/Charge-deficiency-charge-transport-and-comparison-of-dimensions/cmp/1104254604.full, roughly stated as

$$textbf{Proposition}.$$ Let $$K$$ be trace class with integral kernel $$K(vec{x},vec{y})$$ jointly continuous in both components way from a finite set of points, s.t. $$K(vec{x},vec{x})in L^1$$ in neighbourhoods of these points, then
$$text{Tr},K = int_{mathbb{R}^2}K(vec{x},vec{x} ),dvec{x},.$$

Is this proposition useful in any way?

## Functional equation and/or growth estimates for a shifted L function

Consider the $$L$$-series defined by
$$L_{alpha,chi}(s) = sum_{ngeq 1} frac{e^{2pi i alpha Omega(n)} chi(n)}{n^s} = prod_p left(1 – frac{e^{2pi i alpha} chi(p)}{p^s}right)^{-1}.$$
It should have an analytic continuation the left of $$Re s = 1$$.
Does it have any poles other than (possibly) $$s=1$$? Does it satisfy a functional equation? Or rather (since this is why I want a functional equation): what kind of growth estimates do we have on $$L_{alpha,chi}(sigma + i t)$$ for fixed $$sigma < 1$$ and variable $$t$$, $$|t|to infty$$?

(I’m sure this function is known and has most likely been studied – I just do not remember where I have ever seen it.)

## functional analysis – The normed linear space \$l_p\$ ,\$1leq p

The normed linear space $$l_p$$ ,$$1leq p is separable.

Proof:

The normed linear space $$l_p$$ ,$$1leq p is separable. The set $$S$$ of all sequence having finitely many non zero rationals hence $$S$$ is countable.

Now we claim that $$S$$ is dense subset of $$X$$.

Let $$(x_n)_{ngeq 1}=xin l^p,1leq p then $$sum_{i=1}^{infty}|x_i|^p. Thus this series is convergent hence tail of this series is also convergent i.e., for given $$epsilon>0$$, there exist $$ninBbb{N}$$ such that,
$$sum_{i=N}^{infty}|x_i|^p

let $$y_1,y_2,…,y_n$$ be non zero rational coordinates, Since rationals are dense in $$Bbb{R}$$, therefore for given $$epsilon>0$$,

$$|x_i-y_i|<(frac{1}{2n})^pepsilon.$$
Therefore,
$$sum_{i=1}^{N-1}|x_i-y_i|^p

Now $$d(x,y)=||x-y||=(sum_{i=1}^infty|x_i-y_i|^p)^frac{1}{p}=(sum_{i=1}^{N-1}|x_i-y_i|^p+sum_{i=N}^{infty}|x_i|^p)^frac{1}{p}<(frac{epsilon^p}{2}+frac{epsilon^p}{2})^frac{1}{p}=epsilon.$$ Hence any open ball containing $$x$$ contains some points of $$S$$, thus $$S$$ is dense in $$l_p$$ space . This completes the proof.

Is this proof correct?

## functional programming – Passing Multiple Arguments Including A List To A Function In Haskell

Your use of `mutliplesOfKLessThanN` is not correct

``````mutliplesOfKLessThanN((3,5) 1000)
``````

Is not interpreted by Haskell as

Apply `mutliplesOfKLessThanN` with `(3,5)` and `1000`.

but instead it is interpteted as

Apply `(3,5)` to `1000` and apply `multiplesOfKLessThanN` to the result.

I think your misconception is in how function application occurs. In many languages function application requires parentheses e.g. `f(x)`. For Haskell parentheses, only ever mean do this operation first, and function application is achieved by putting things next to each other. So `f(x)` works in Haskell because it is the same as `f x`, but `f(x y)` is the same as `f(x(y))` and tells Haskell to evaluate `x y` first and then give it to `f`.

With your code Haskell can’t apply `(3,5)` as a function, which is what Haskell is telling you, it expected a function (in fact a specific type of function).

The proper way to write this would be

``````multiplesOfKLessThanN (3,5) 1000
``````

This should handle that main error you are getting.

## reactjs – How to pass data between two functional component siblings through Router Route?

I am still learning React and I was stuck with this issue.
I need to pass props from parent to two children as this is the way the siblings can communicate with eachother.

I tried to use render to pass the props, but it does not work.

Here the parent component which has cart, setCart, cartCopy and setCartCopy, which is what I need to pass to the children.

``````function App() {
const (cart, setCart) = useState(());
const (cartCopy, setCartCopy) = useState(())

return (
<div >
<img className='logo' src={logo} alt='NB' />
<Router>
<Nav />
<Route exact path="/" component={Home} />
<Route exact path="/products" component={Products} render={(props) => <Products {...props} cart={cart} setCart={setCart} cartCopy={cartCopy} setCartCopy={setCartCopy} />} />
<Route exact path="/cart" component={Cart} render={(props) => <Cart {...props} cart={cart} setCart={setCart} cartCopy={cartCopy} setCartCopy={setCartCopy} />} />
<Route exact path="/signin" component={Signin} />
<Route exact path="/register" component={Register} />
</Router>
</div>

);
}
``````

Here one child component, which needs to recieve from parent in order to use the data:

``````const Cart = ({ cart, setCart, cartCopy, setCartCopy }) => {

const clearCartItem = (itemID) => {

let cartCopy = (...cart)
cartCopy = cartCopy.filter(item => item._id != itemID);

setCart(cartCopy);
let cartString = JSON.stringify(cartCopy)
localStorage.setItem('cart', cartString)
setCartCopy(cartCopy)
}
``````

Thank you!

## functional analysis – Trying to understand definition of Lie ideal for C*-algebras

Let $$A$$ be a $$C^*$$-algebra. A sub space $$I$$ of $$A$$ is called Lie ideal of A if $$[I,A]= IA-AI subset I$$

Since I contains $$0$$, isn’t it this definition equivalent to definition of two sided ideal of $$C^*$$-algerba?

Most probably I’m missing something in the definition of Lie ideal. Any ideas?

## functional analysis – Understanding the expoenetial operator

Let $$A$$ a bounded operator on a Hilbert space. Define the exponential operator by:
$$begin{eqnarray*} e^{A}=sum_{n=0}^{infty}frac{A^{n}}{n!} (A^{0}=I) end{eqnarray*}$$

I undesrtand this exponential operator like this: If $$xin H$$ then:
$$begin{eqnarray*} e^{A}(x)= sum_{n=0}^{infty} frac{A^{n}(x)}{n!} (A^{0}(x)=x) end{eqnarray*}$$
But if I think in for example $$(e^{A})^{2}$$. How can I see this function? I understand $$(e^{A})^{2}$$ by the composition of $$e^{A}$$ with itself.

## What is the granularity of specifying applicative or normal evaluation order in lambda calculus (and functional languages)?

In lambda calculus and other functional languages, what is the granularity of specifying applicative or normal evaluation order:

• language-wide (for all the function calls),
• individual functions (for all the function calls for a given function),
• individual function calls?

I also seem to remember in some functional language (Haskell?), the evaluation order is a language wide choice. By default, it is either applicative or normal order, and it can be switched by a statement in programs. When it is switched, will the evaluation order for all the functions defined both before and after the switch be changed? Can we switch the evaluation order for all the calls to a function, without affecting the calls to other functions? Can we switch the evaluation order of a function call, without affecting other function calls?

I have the above question, because for the normal order sequencing combinator `Seq`, when evaluating `((Seq (display “hello”)) (display “world”))` to `(λz.(display “world”) (display “hello”)`, I am wondering if `(λz.(display “world”) (display “hello”)` is evaluated in applicative or normal order, since I guess both “hello” and “world” are supposed to be printed but in normal order “hello” isn’t printed.

Thanks.