Other exercises on $L^p$ spaces
1) Let $u \in L^1(0,+\infty)$; suppose that there exists $L=\lim_{t \to +\infty}u(t)$. Prove that $L=0$.
2) Let $u \in L^1(0,+\infty)$; prove that if $u$ is uniformly continuous then $\lim_{t \to +\infty}u(t)=0$.
3) Find $u \in L^1(0,+\infty)$ continuous in such a way that $\lim_{t \to +\infty}u(t)$ does not exist.
2) Let $u \in L^1(0,+\infty)$; prove that if $u$ is uniformly continuous then $\lim_{t \to +\infty}u(t)=0$.
3) Find $u \in L^1(0,+\infty)$ continuous in such a way that $\lim_{t \to +\infty}u(t)$ does not exist.
Risposte
"ubermensch":
In the first exercice there isn't the hypothesis that $u$ is continuos ...
But there is the hypothesis that the limit exists as confirmed by L uca
[size=75]killed[/size]
But in the first exercise the existence of the limit was assumed....
"ubermensch":
In the first exercice there isn't the hypothesis that $u$ is continuos ...
Yeah! My answer was about my answers. I was happy not to be guilty
In the first exercice there isn't the hypothesis that $u$ is continuos ...
Let me notice that I answered questions 2) and 3), where it is said that the functions are continuous (representatives of the equivalence class...).
In italiano: ho risposto a 2) e 3), dove si assume che le funzioni siano continue.
[size=75]BANG! ZIP (mancato, accidenti!) BANG! ZIIING (e che mira scadente!!! Ma no, è caricata a salve, per la miseria)[/size]
In italiano: ho risposto a 2) e 3), dove si assume che le funzioni siano continue.
[size=75]BANG! ZIP (mancato, accidenti!) BANG! ZIIING (e che mira scadente!!! Ma no, è caricata a salve, per la miseria)[/size]
Scusate parlo un attimo in italiano: ma il limite di una funzione $L^p$ si può considerare in maniera così spensierata? Chiarisco: potendo variare i valori della funzione su un insieme di misura nulla, non mi pare che abbia un senso ovvio il calcolo del limite (essendo di fatto una valutazione puntuale della funzione)... Attendo numi! scusate l'italiano. Non uccidetemi.
3) are the usual spikes
2) is really nice. Never seen it. But you "feel" that it must be true. How to convert feeling into a proof? I would use the "limsup". Assume is positive. Then I find a sequence going to $+oo$ where f is greater that $\alpha$. Using uniform continuity, given $\epsilon = \alpha/2$, I find a denumerable set of intervals (I can make them disjoint! Can't I?) of length $\delta$ where $u$ is greater that $\alpha/2$. I have omitted details, with the hope that they are really such
2) is really nice. Never seen it. But you "feel" that it must be true. How to convert feeling into a proof? I would use the "limsup". Assume is positive. Then I find a sequence going to $+oo$ where f is greater that $\alpha$. Using uniform continuity, given $\epsilon = \alpha/2$, I find a denumerable set of intervals (I can make them disjoint! Can't I?) of length $\delta$ where $u$ is greater that $\alpha/2$. I have omitted details, with the hope that they are really such
Let assume by absurd that $lim_(x rarr +oo)|u(x)|= L > 0 $.
Then for all $x $ large enough, let say for $ x > x_L$ , we have : $|u(x)| >L/2 $.
For such values of $x$ we have also :
$ int_0^x |u(t)|dt= int_0^(x_L)|u(t)|dt+ int_(x_L)^x|u(t)| dt > int_0^(x_L) |u(t)|dt+(L/2)(x-x_L) $.
Then for $ x rarr +oo$ the first member has a finite value, while $(L/2)(x-x_L) $ diverges to $oo$.This is a contradiction.
Therefore the hypothesis $ L > 0 $ was wrong and the only possibility is that, if the limit exists, $L = 0 $ .
Then for all $x $ large enough, let say for $ x > x_L$ , we have : $|u(x)| >L/2 $.
For such values of $x$ we have also :
$ int_0^x |u(t)|dt= int_0^(x_L)|u(t)|dt+ int_(x_L)^x|u(t)| dt > int_0^(x_L) |u(t)|dt+(L/2)(x-x_L) $.
Then for $ x rarr +oo$ the first member has a finite value, while $(L/2)(x-x_L) $ diverges to $oo$.This is a contradiction.
Therefore the hypothesis $ L > 0 $ was wrong and the only possibility is that, if the limit exists, $L = 0 $ .
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