$L^p$ convergence of a sequence of functions
$begingroup$
I'd like a check for the following exercise I found online
Discuss the $L^{p}(0,1)$, $pin[1,+infty]$ convergence of the
sequence of functions $$ u_n(x)= left{ begin{array}{c} n^{4/3}x,
quad 0 < x <frac2n \ 0, quad frac2n leq x <1 \ end{array}
right. $$
Solution
The pointwise limit is $0$:
$lim_{n rightarrow +infty} u_{n}(x)=0$, where $u_n(x)= mathbb{1}_{(0,frac2n)}(x) n^{frac43} x$
In fact, $forall epsilon>0 exists n_0 text{ s.t } forall ngeq n_0:|frac2 n| < epsilon $
I'll study separately the cases for $p=+infty$, and $pgeq 1$.
- $p=+infty$
$lim_n sup_{x in (0,1)} |u_n(x)|=lim_n sup_{x in (0,frac2n)} n^{frac43}x=lim_n 2 n^{frac13} rightarrow +infty$ and thus there's no $L^{infty}$ convergence in $(0,1)$
- $p geq 1$
The integral to be computed is $int_0^1 |u_n(x)|^pdx=frac{2^{p+1}}{p+1} cdot n^{frac{p-3}{3} }$ and taking the limit for $n rightarrow +infty$ this is $0$ iff $frac{p-3}{3}<1$, thus we have $L^{p}(0,1)$ convergence for $p in [1,6)$
$square$
functional-analysis convergence
asked Dec 11 '18 at 22:25
VoBVoB
734513
$endgroup$
add a comment |
$begingroup$
I'd like a check for the following exercise I found online
Discuss the $L^{p}(0,1)$, $pin[1,+infty]$ convergence of the
sequence of functions $$ u_n(x)= left{ begin{array}{c} n^{4/3}x,
quad 0 < x <frac2n \ 0, quad frac2n leq x <1 \ end{array}
right. $$
Solution
The pointwise limit is $0$:
$lim_{n rightarrow +infty} u_{n}(x)=0$, where $u_n(x)= mathbb{1}_{(0,frac2n)}(x) n^{frac43} x$
In fact, $forall epsilon>0 exists n_0 text{ s.t } forall ngeq n_0:|frac2 n| < epsilon $
I'll study separately the cases for $p=+infty$, and $pgeq 1$.
- $p=+infty$
$lim_n sup_{x in (0,1)} |u_n(x)|=lim_n sup_{x in (0,frac2n)} n^{frac43}x=lim_n 2 n^{frac13} rightarrow +infty$ and thus there's no $L^{infty}$ convergence in $(0,1)$
- $p geq 1$
The integral to be computed is $int_0^1 |u_n(x)|^pdx=frac{2^{p+1}}{p+1} cdot n^{frac{p-3}{3} }$ and taking the limit for $n rightarrow +infty$ this is $0$ iff $frac{p-3}{3}<1$, thus we have $L^{p}(0,1)$ convergence for $p in [1,6)$
$square$
functional-analysis convergence
asked Dec 11 '18 at 22:25
VoBVoB
734513
$endgroup$
add a comment |
$begingroup$
I'd like a check for the following exercise I found online
Discuss the $L^{p}(0,1)$, $pin[1,+infty]$ convergence of the
sequence of functions $$ u_n(x)= left{ begin{array}{c} n^{4/3}x,
quad 0 < x <frac2n \ 0, quad frac2n leq x <1 \ end{array}
right. $$
Solution
The pointwise limit is $0$:
$lim_{n rightarrow +infty} u_{n}(x)=0$, where $u_n(x)= mathbb{1}_{(0,frac2n)}(x) n^{frac43} x$
In fact, $forall epsilon>0 exists n_0 text{ s.t } forall ngeq n_0:|frac2 n| < epsilon $
I'll study separately the cases for $p=+infty$, and $pgeq 1$.
- $p=+infty$
$lim_n sup_{x in (0,1)} |u_n(x)|=lim_n sup_{x in (0,frac2n)} n^{frac43}x=lim_n 2 n^{frac13} rightarrow +infty$ and thus there's no $L^{infty}$ convergence in $(0,1)$
- $p geq 1$
The integral to be computed is $int_0^1 |u_n(x)|^pdx=frac{2^{p+1}}{p+1} cdot n^{frac{p-3}{3} }$ and taking the limit for $n rightarrow +infty$ this is $0$ iff $frac{p-3}{3}<1$, thus we have $L^{p}(0,1)$ convergence for $p in [1,6)$
$square$
functional-analysis convergence
asked Dec 11 '18 at 22:25
VoBVoB
734513
$endgroup$
I'd like a check for the following exercise I found online
Discuss the $L^{p}(0,1)$, $pin[1,+infty]$ convergence of the
sequence of functions $$ u_n(x)= left{ begin{array}{c} n^{4/3}x,
quad 0 < x <frac2n \ 0, quad frac2n leq x <1 \ end{array}
right. $$
Solution
The pointwise limit is $0$:
$lim_{n rightarrow +infty} u_{n}(x)=0$, where $u_n(x)= mathbb{1}_{(0,frac2n)}(x) n^{frac43} x$
In fact, $forall epsilon>0 exists n_0 text{ s.t } forall ngeq n_0:|frac2 n| < epsilon $
I'll study separately the cases for $p=+infty$, and $pgeq 1$.
- $p=+infty$
$lim_n sup_{x in (0,1)} |u_n(x)|=lim_n sup_{x in (0,frac2n)} n^{frac43}x=lim_n 2 n^{frac13} rightarrow +infty$ and thus there's no $L^{infty}$ convergence in $(0,1)$
- $p geq 1$
The integral to be computed is $int_0^1 |u_n(x)|^pdx=frac{2^{p+1}}{p+1} cdot n^{frac{p-3}{3} }$ and taking the limit for $n rightarrow +infty$ this is $0$ iff $frac{p-3}{3}<1$, thus we have $L^{p}(0,1)$ convergence for $p in [1,6)$
$square$
functional-analysis convergence
functional-analysis convergence
asked Dec 11 '18 at 22:25
VoBVoB
734513
asked Dec 11 '18 at 22:25
VoBVoB
734513
asked Dec 11 '18 at 22:25
VoBVoB
734513
asked Dec 11 '18 at 22:25
VoBVoB
734513
asked Dec 11 '18 at 22:25
VoBVoB
734513
734513
add a comment |
add a comment |
1 Answer
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You have made two mistakes. $int u_n^{p}= cn^{4p/3} frac 1 {n^{p+1}}$ for some positive constant $c$ and this tends to $0$ iff $frac {4p} 3 <p+1$ and this is true iff $p < 3$.
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
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1 Answer
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1 Answer
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$begingroup$
You have made two mistakes. $int u_n^{p}= cn^{4p/3} frac 1 {n^{p+1}}$ for some positive constant $c$ and this tends to $0$ iff $frac {4p} 3 <p+1$ and this is true iff $p < 3$.
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
add a comment |
$begingroup$
You have made two mistakes. $int u_n^{p}= cn^{4p/3} frac 1 {n^{p+1}}$ for some positive constant $c$ and this tends to $0$ iff $frac {4p} 3 <p+1$ and this is true iff $p < 3$.
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
add a comment |
$begingroup$
You have made two mistakes. $int u_n^{p}= cn^{4p/3} frac 1 {n^{p+1}}$ for some positive constant $c$ and this tends to $0$ iff $frac {4p} 3 <p+1$ and this is true iff $p < 3$.
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
$endgroup$
You have made two mistakes. $int u_n^{p}= cn^{4p/3} frac 1 {n^{p+1}}$ for some positive constant $c$ and this tends to $0$ iff $frac {4p} 3 <p+1$ and this is true iff $p < 3$.
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
answered Dec 11 '18 at 23:42
Kavi Rama MurthyKavi Rama Murthy
61.8k42262
61.8k42262
add a comment |
add a comment |
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