MHB Norm in R^d with Lebesgue measure

[mathmari]

Hey!

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)

 

[Opalg]

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.

 

[mathmari]

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)

 

[Euge]

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

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

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

 

[mathmari]

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

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

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)

 

[Euge]

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 $y\in \Bbb R^d$,

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

Therefore, for all $y\in \Bbb R^d$, $f_y\in L^p$ with $\|f_y\|_p = \|f\|_p$.

 

[mathmari]

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.

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

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

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

Could you explain it further to me?? (Wondering)

 

[Euge]

Which part do you need more explanation, mathmari?

 

[mathmari]

Which part do you need more explanation, mathmari?

Why do we have to show it first for continuous functions with compact support??

 

[Euge]

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.