Norm in R^d with Lebesgue measure

In summary, the norm in R^d with Lebesgue measure is a mathematical concept used to measure the size or length of a vector in d-dimensional space. It is calculated by taking the square root of the sum of the squares of the components of the vector and has properties such as non-negativity, homogeneity, and the triangle inequality. This concept is fundamental in linear algebra and functional analysis, with applications in optimization, statistics, and physics. It is also closely related to the Lebesgue measure, as it is used to measure the "size" of vectors in d-dimensional space.
  • #1
mathmari
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Hey! :eek:

In $\mathbb{R}^d$ with the Lebesgue measure if $f \in L^p, 1 \leq p < +\infty$, and if for each $y$ we set $f_y(x)=f(x+y)$ then:
  1. $f_y \in L^p$ and $||f||_p=||f_y||_p$
  2. $\lim_{y \rightarrow 0} ||f-f_y||_p=0$

Could you give me some hints how to show that?? (Wondering)
 
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  • #2
1. follows from the fact that Lebesgue measure is translation-invariant.

For 2., I think that you may need to start by proving this for some class of functions that is dense in $L^p$. For example, can you show that it holds for continuous functions with compact support, or alternatively for simple functions in $L^p$? Then deduce that the result holds for all $L^p$ functions.
 
  • #3
Opalg said:
1. follows from the fact that Lebesgue measure is translation-invariant.

For 2., I think that you may need to start by proving this for some class of functions that is dense in $L^p$. For example, can you show that it holds for continuous functions with compact support, or alternatively for simple functions in $L^p$? Then deduce that the result holds for all $L^p$ functions.

  1. $$L_p=\{f: X \rightarrow \overline{\mathbb{R}} \text{ or } \mathbb{C} : f \text{ measurable and } \int |f|^pd \mu<+\infty\}$$

    We have that $f \in L^p$, so $\int |f|^p d \mu < +\infty$.

    $\int |f_y|^pd \mu=\int |f(x+y)|^pd \mu \ \ \ \ \ \overset{\text{ translation-invariant }}{=} \ \ \ \ \ \int |f(x)|^pd \mu < +\infty$

    So, $\int |f_y|^p d \mu<+\infty$ that means that $f_y \in L^p$.

    Is this correct?? (Wondering)$||f||_p= \left ( \int |f|^p d \mu \right )^{1/p} \ \ \ \ \ \overset{\text{ translation-invariant }}{=} \ \ \ \ \ \left ( \int |f(x+y)|^p d \mu \right )^{1/p}=\left ( \int |f_y|^p d \mu \right )^{1/p}=||f_y||_p$

    Is this right?? (Wondering)
  2. $||f-f_y||_p \ \ \ \ \ \overset{\text{ Minkowsi }}{ \leq } \ \ \ \ \ ||f||_p+||f_y||_p=||f||_p+||f||_p=2||f||_p$

    Can we use this?? Or isn't it correct?? (Wondering)
 
  • #4
As Opalg suggested, in part 2, first prove the result when $f$ is continuous with compact support. In the general case, given $\varepsilon > 0$, there exists a continuous function $g$ with compact support such that $||f - g||_p < \frac{\varepsilon}{3}$. Choose $\delta > 0$ such that $||g - g_y||_p < \frac{\varepsilon}{3}$ whenever $|y| < \delta$. By translation invariance of the Lebesgue measure, $\|g_y - f_y\|_p = \|g - f\|_p < \frac{\varepsilon}{3}$ for all $y$. Thus

\(\displaystyle \|f - f_y\|_p \le \|f - g\|_p + \|g - g_y\|_p + \|g_y - f_y\|_p < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon\)

for all $y$ with $|y| < \delta$. Since $\varepsilon$ was arbitrary, the result follows.
 
  • #5
Euge said:
As Opalg suggested, in part 2, first prove the result when $f$ is continuous with compact support. In the general case, given $\varepsilon > 0$, there exists a continuous function $g$ with compact support such that $||f - g||_p < \frac{\varepsilon}{3}$. Choose $\delta > 0$ such that $||g - g_y||_p < \frac{\varepsilon}{3}$ whenever $|y| < \delta$. By translation invariance of the Lebesgue measure, $\|g_y - f_y\|_p = \|g - f\|_p < \frac{\varepsilon}{3}$ for all $y$. Thus

\(\displaystyle \|f - f_y\|_p \le \|f - g\|_p + \|g - g_y\|_p + \|g_y - f_y\|_p < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon\)

for all $y$ with $|y| < \delta$. Since $\varepsilon$ was arbitrary, the result follows.

Ok, I will think about it! (Wait)

Is my proof of the first sentence correct?? (Wondering)
 
  • #6
mathmari said:
Is my proof of the first sentence correct?? (Wondering)

What you have is almost correct, but it's not a proof yet. Here's an example of a proof of part 1:

For each $y\in \Bbb R^d$, $f_y$ is the composition of the measurable function $f$ and the continuous map $x \mapsto x + y$ from $\Bbb R^d$ to $\Bbb R^d$. Hence $f_y$ is measurable for every $y$. By translational invariance of the Lebesgue measure, for all $y\in \Bbb R^d$,

\(\displaystyle \|f_y\|_p^p = \int_{\Bbb R^d}|f(x - y)|^p\, dx = \int_{\Bbb R^d} |f(x)|^p\, dx = \|f\|_p^p < \infty\)

Therefore, for all $y\in \Bbb R^d$, $f_y\in L^p$ with $\|f_y\|_p = \|f\|_p$.
 
  • #7
Opalg said:
For 2., I think that you may need to start by proving this for some class of functions that is dense in $L^p$. For example, can you show that it holds for continuous functions with compact support, or alternatively for simple functions in $L^p$? Then deduce that the result holds for all $L^p$ functions.

Euge said:
As Opalg suggested, in part 2, first prove the result when $f$ is continuous with compact support. In the general case, given $\varepsilon > 0$, there exists a continuous function $g$ with compact support such that $||f - g||_p < \frac{\varepsilon}{3}$. Choose $\delta > 0$ such that $||g - g_y||_p < \frac{\varepsilon}{3}$ whenever $|y| < \delta$. By translation invariance of the Lebesgue measure, $\|g_y - f_y\|_p = \|g - f\|_p < \frac{\varepsilon}{3}$ for all $y$. Thus

\(\displaystyle \|f - f_y\|_p \le \|f - g\|_p + \|g - g_y\|_p + \|g_y - f_y\|_p < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon\)

for all $y$ with $|y| < \delta$. Since $\varepsilon$ was arbitrary, the result follows.

Could you explain it further to me?? (Wondering)
 
  • #8
Which part do you need more explanation, mathmari?
 
  • #9
Euge said:
Which part do you need more explanation, mathmari?

Why do we have to show it first for continuous functions with compact support??
 
  • #10
It doesn't have to be (look at Opalg's suggestions), but it makes things easier to find estimates for $\|f - f_y\|_p$ for compactly supported continuous functions $f$, as you have seen. Once the result is proven for this class of functions, then we can prove the general case by the argument I showed before or by application of Fatou's lemma. These are common methods of proof when dealing with convergence theorems involving $L^p$ norms.
 

FAQ: Norm in R^d with Lebesgue measure

1.

What is the norm in R^d with Lebesgue measure?

The norm in R^d with Lebesgue measure is a mathematical concept used to measure the size or length of a vector in d-dimensional space. It is usually denoted by ||x|| and is defined as the square root of the sum of the squares of the components of the vector.

2.

How is the norm in R^d with Lebesgue measure calculated?

The norm in R^d with Lebesgue measure is calculated by taking the square root of the sum of the squares of the components of a vector in d-dimensional space. Mathematically, it can be represented as ||x|| = √(x1^2 + x2^2 + ... + xd^2).

3.

What are the properties of the norm in R^d with Lebesgue measure?

The norm in R^d with Lebesgue measure has several properties, including non-negativity, homogeneity, and the triangle inequality. Non-negativity means that the norm is always greater than or equal to zero. Homogeneity means that multiplying a vector by a constant also multiplies its norm by the same constant. The triangle inequality states that the norm of the sum of two vectors is always less than or equal to the sum of their individual norms.

4.

What is the significance of the norm in R^d with Lebesgue measure in mathematics?

The norm in R^d with Lebesgue measure is a fundamental concept in linear algebra and functional analysis. It is used to define the distance between two vectors, which is essential in many applications, such as optimization, statistics, and physics. It also plays a crucial role in defining inner products and orthogonality in vector spaces.

5.

How is the norm in R^d with Lebesgue measure related to the Lebesgue measure?

The Lebesgue measure is a mathematical concept used to assign a size or volume to subsets of d-dimensional space. The norm in R^d with Lebesgue measure is closely related to the Lebesgue measure because it is used to measure the "size" of vectors in d-dimensional space. In fact, the Lebesgue measure of a set can be calculated by integrating the norm function over that set.

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