Parametric function
Determine fo which values of the parameter $alpha in RR $, the function $ g(x) = (e^(-1/x)*sin(x^2+1))/x^alpha $ belongs to $L^p (0,+oo) $ .
Risposte
Since I have seen no replies I will indicate herebelow the solution :
The $ g(x)$ function is continuous in $(0,+oo)$ [independently from $alpha $ value] and it is therefore measurable.
Since $lim_(x rarr 0^(+))g(x) = 0 $ independently from $ alpha $ value,then $g(x) in L^oo([0,M]), AA M >0 ; $ the function can be extended in $0$ to a continuous function in the compact $ [0,M ]$.
We have to find for which values of $alpha$ , $ g in L^p(1,+oo)$.
Let's consider first $p=oo$.
Since $ g(x) $ is continuous in $[1,+oo)$, $ g in L^p(1,+oo) $ iff $lim_(x rarr +oo)"sup" g(x) <+oo$ .
We note that if $ alpha >0$ , then $lim_(x rarr +oo)g(x)=0$,while if $alpha < 0 $ then $lim_(x rarr +oo)g(x)= +oo$ and if $alpha=0$ then $lim_(x rarr+oo)"sup" g(x)=1$.
In conclusion $g(x) in L^(oo)(0,+oo)$ iff $alpha>=0$.
Let's now consider $p in RR$.Since $sin(x^2+1)$ is limited in module in $(0,+oo)$, then $ g in L^p(0,+oo)$ for the values of $alpha $ such that $1/x^(alpha*p) in L^(1)(0,+oo)$, that is : $alpha >1/p$.
Therefore $g(x) inL^(oo)(0,+oo)$ if $ alpha >=0$ ;$g(x) in L^p(0,+oo)$ if $alpha >1/p$ ( it can be shown that this condition is necessary and sufficient for $g in L^p(0,+oo)$.).
The $ g(x)$ function is continuous in $(0,+oo)$ [independently from $alpha $ value] and it is therefore measurable.
Since $lim_(x rarr 0^(+))g(x) = 0 $ independently from $ alpha $ value,then $g(x) in L^oo([0,M]), AA M >0 ; $ the function can be extended in $0$ to a continuous function in the compact $ [0,M ]$.
We have to find for which values of $alpha$ , $ g in L^p(1,+oo)$.
Let's consider first $p=oo$.
Since $ g(x) $ is continuous in $[1,+oo)$, $ g in L^p(1,+oo) $ iff $lim_(x rarr +oo)"sup" g(x) <+oo$ .
We note that if $ alpha >0$ , then $lim_(x rarr +oo)g(x)=0$,while if $alpha < 0 $ then $lim_(x rarr +oo)g(x)= +oo$ and if $alpha=0$ then $lim_(x rarr+oo)"sup" g(x)=1$.
In conclusion $g(x) in L^(oo)(0,+oo)$ iff $alpha>=0$.
Let's now consider $p in RR$.Since $sin(x^2+1)$ is limited in module in $(0,+oo)$, then $ g in L^p(0,+oo)$ for the values of $alpha $ such that $1/x^(alpha*p) in L^(1)(0,+oo)$, that is : $alpha >1/p$.
Therefore $g(x) inL^(oo)(0,+oo)$ if $ alpha >=0$ ;$g(x) in L^p(0,+oo)$ if $alpha >1/p$ ( it can be shown that this condition is necessary and sufficient for $g in L^p(0,+oo)$.).
It's not difficult....
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