Two limits
Let $f in L^p(RR^n)$, $1<=p
$lim_(||h|| to oo) int|f(x+h)-f(x)|^p dx$.
$lim_(||h|| to oo) int|f(x+h)-f(x)|^p dx$.
Risposte
I had completely forgotten this problem, and now I discover I had made a mistake: $f in L^p(mathbb(R)^n)$.
(I hope it hasn't influenced too much, since the error was quite easy to discover).
I) Given $epsilon > 0,$ we can find a continuous, compactly supported function $g$ such that $||f - g||_p < epsilon$.
$g$ is uniformly continuous, hence $g(x + h)$ converges uniformly to $g(x)$ as $hto0,$ and the integral
can be taken over a set of finite measure. Thus the limit is $0$.
II) Approximate $f$ in $L^p$ norm by $g$ which has compact support. For large enough $||h||,$ $g(x)$ and $g(x + h)$
have disjoint support. Let $A$ be a bounded set which contains the support of $g$. Then
$int_{mathbb{R}^n}|g(x + h) - g(x)|^p dx = int_{A - h}|g(x + h)|^p dx + int_A|g(x)|^p dx= 2int_A|g(x)|^p dx = 2||g||_p^p$.
Now we can use the fact that $g$ approximates $f$ and conclude that the limit is $2||f||_p^p$.
(I hope it hasn't influenced too much, since the error was quite easy to discover).
I) Given $epsilon > 0,$ we can find a continuous, compactly supported function $g$ such that $||f - g||_p < epsilon$.
$g$ is uniformly continuous, hence $g(x + h)$ converges uniformly to $g(x)$ as $hto0,$ and the integral
can be taken over a set of finite measure. Thus the limit is $0$.
II) Approximate $f$ in $L^p$ norm by $g$ which has compact support. For large enough $||h||,$ $g(x)$ and $g(x + h)$
have disjoint support. Let $A$ be a bounded set which contains the support of $g$. Then
$int_{mathbb{R}^n}|g(x + h) - g(x)|^p dx = int_{A - h}|g(x + h)|^p dx + int_A|g(x)|^p dx= 2int_A|g(x)|^p dx = 2||g||_p^p$.
Now we can use the fact that $g$ approximates $f$ and conclude that the limit is $2||f||_p^p$.
I am still interested in the solution 
Interesting exercises , but so far...
A hint would help .
A hint would help .
Accedi a tutti gli appunti
Tutor AI: studia meglio e in meno tempo