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If $g\in L^p(\mathbb{R}^n)$ and $1\leq p<\infty$ then show $\lim_{|t|\to \infty}\lVert g_{(t)}+g\rVert_p=2^{1/p}\lVert g\rVert_p,$

where $g_{(t)}(x):=g(t+x)$.

Any hints? Try to give me only hints/outlines not complete solutions

Not sure where to go from there?

  • 0
    You can try to prove it for characteristic function of hypercube , and then for characteristic function of any set, with using iterated limits, starting from t. An example of using iterated limits is here: http://math.stackexchange.com/questions/677108/proof-of-riemann-lebesgue-lemma/711617#711617 ,you can also consider inequality described here: http://math.stackexchange.com/questions/458230/continuity-of-l1-functions-with-respect-to-translation/458235#458235, or some tricks from this document: http://fractal.math.unr.edu/~ejolson/761/notes/761sep12.pdf2014-04-23

3 Answers 3

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Hint: Prove this for compactly supported functions.

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    said: This might be because you forgot that the $L^p$ norm is the power $1/p$ of an integral. Hence the factor $2^{1/p}$ is only natural.2012-10-29
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Hint:

  1. First show it for characteristic functions $\chi_I$ where $I$ is some interval.

  2. Using 1. prove this for simple functions (i.e. finite sums $\sum \alpha_I\chi_I$).

  3. Prove the general statement by approximating a general function $g\in L^p$ by those of 2.

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Hint:

Let's introduce $B_r(P)$ as it is defined and $\|\ f\|_{L_{p}{\{\text{Area of integrating}\}}}=\left(\int\limits_{\text{Area of integrating}}|f(x)|^p dx\right)^{\frac{1}{p}}$. The key to solving the problem are triangle inequalities as it is shown below: $\|g\|_{L_{p}\{B_{0.5|t|}(0)\}}\leqslant\|g_{(t)}+g\|_{L_{p}\{B_{0.5|t|}(0)\}}+\|g_{(t)}\|_{L_{p}\{B_{0.5|t|}(0)\}}$

and: $\|g_{(t)}+g\|_{L_{p}\{B_{0.5|t|}(0)\}}\leqslant \|g\|_{L_{p}\{B_{0.5|t|}(0)\}}+\|g_{(t)}\|_{L_{p}\{B_{0.5|t|}(0)\}}$

Notice that: $\|g_{(t)}+g\|_{L_p}=\left(\|g_{(t)}+g\|_{L_{p}\{B_{0.5|t|}(0)\}}^p+\|g_{(t)}+g\|_{L_{p}\{B_{0.5|t|}(-t)\}}^p+\|g_{(t)}+g\|_{L_{p}\{\overline{B_{0.5|t|}(0)\cup B_{0.5|t|}(-t)}\}}^p\right)^{\frac{1}{p}}$