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henrikloev · **Joined:** Sat, 8 Oct 2011 17:57:59 UTC **Posts:** 5

Hello,

Inside two measure spaces: (X, sigmaAlgebraA, b), (Y, sigmaAlgebraB, v)
Suppose you have a function u : X x Y -> [0,infinity]

What does the following mean in this setting?

x |-> u(x, y)

or

x |-> Integral ( u(x, y)v(dy)

I dont understand the meaning of the symbol '|->'. Tried to google it several times with no luck.

In Latex its called /mapsto.

Thank you for you help.

Henrik Løv

Shadow · **Posted:** Wed, 12 Oct 2011 19:38:55 UTC

henrikloev wrote:

Hello,

Inside two measure spaces: (X, sigmaAlgebraA, b), (Y, sigmaAlgebraB, v)
Suppose you have a function u : X x Y -> [0,infinity]

What does the following mean in this setting?

x |-> u(x, y)

or

x |-> Integral ( u(x, y)v(dy)

I dont understand the meaning of the symbol '|->'. Tried to google it several times with no luck.

In Latex its called /mapsto.

Thank you for you help.

Henrik Løv

It means you have a function which takes x and gives you the function $u(x,-): Y \to [0,\infty]$ .

Basically the $x \mapsto$ notation means you're defining the function by telling you what the image of a point

is.

henrikloev · **Joined:** Sat, 8 Oct 2011 17:57:59 UTC **Posts:** 5

Thank you for your answer.

So,

$x \mapsto u(x, y)$ = u(x_0, y)

?

I dont fully understand what you mean by u(x, -)

Shadow · **Posted:** Wed, 12 Oct 2011 23:21:55 UTC

henrikloev wrote:

Thank you for your answer.

So,

$x \mapsto u(x, y)$ = u(x_0, y)

?

I dont fully understand what you mean by u(x, -)

is a function from $Y\to [0\infty]$ which takes the value u(x,y)

at

.

I don't know what your x_0

is btw, so I have no comment on that bit.

henrikloev · **Joined:** Sat, 8 Oct 2011 17:57:59 UTC **Posts:** 5

Suppose

.

What would $x \mapsto u(x, y)$ then be?

I don't understand, because $x \in X$ and the function you write is $Y \to \infty$ ?

outermeasure · **Posted:** Thu, 13 Oct 2011 14:23:42 UTC

henrikloev wrote:

Suppose

.

What would $x \mapsto u(x, y)$ then be?

I don't understand, because $x \in X$ and the function you write is $Y \to \infty$ ?

In this case, $x\mapsto u(x,y)$ would just be the function that takes every x and add y to it.

It is just a notation. $f\colon x\mapsto x^2$ is the function f(x):=x^2

, or if you prefer the lambda-calculus notation, $f(x)=\lambda x.x^2$ . Of course, this notation doesn't tell you the domain or the codomain of the function, which needed to be specified elsewhere, e.g. $f(x)=\lambda x\in A. x^2$ .

Back to $x\mapsto \int_Y u(x,y)\nu(\,\mathrm{d}y)$ --- it is the function that takes whatever x you input, and spits out the value $\int_Y u(x,y)\nu(\,\mathrm{d}y)$ , i.e. the function that takes x in X as input, and output the integral of the nonnegative (measurable) function $u(x,-)\colon Y\to[0,\infty]$ with respect to measure $\nu$ on Y. Note the two different types of arrows here --- $f\colon A\to B$ is used for specifying the domain/codomain, and $f\colon x\mapsto x+y$ is used for specifying the value of the function at (generic) argument.

Shadow · **Posted:** Thu, 13 Oct 2011 17:39:02 UTC

outermeasure wrote:

henrikloev wrote:

Suppose

.

What would $x \mapsto u(x, y)$ then be?

I don't understand, because $x \in X$ and the function you write is $Y \to \infty$ ?

In this case, $x\mapsto u(x,y)$ would just be the function that takes every x and add y to it.

It is just a notation. $f\colon x\mapsto x^2$ is the function f(x):=x^2

, or if you prefer the lambda-calculus notation, $f(x)=\lambda x.x^2$ . Of course, this notation doesn't tell you the domain or the codomain of the function, which needed to be specified elsewhere, e.g. $f(x)=\lambda x\in A. x^2$ .

Back to $x\mapsto \int_Y u(x,y)\nu(\,\mathrm{d}y)$ --- it is the function that takes whatever x you input, and spits out the value $\int_Y u(x,y)\nu(\,\mathrm{d}y)$ , i.e. the function that takes x in X as input, and output the integral of the nonnegative (measurable) function $u(x,-)\colon Y\to[0,\infty]$ with respect to measure $\nu$ on Y. Note the two different types of arrows here --- $f\colon A\to B$ is used for specifying the domain/codomain, and $f\colon x\mapsto x+y$ is used for specifying the value of the function at (generic) argument.

I don't know, I interpreted his question as assigning x to the translation by x operator on

, since he said

was a function. I agree that

does what you said, but we're talking about the assignment $x\mapsto u(x,y)$ .

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