Mixing
Show that $phi(r) = ||f||_{L^r(0,1)}$, $rin [1,2]$ is continuous function of r.
$begingroup$
Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.
Do I need to use the definition of continuity to solve this question? I don't know how to start this question.
real-analysis analysis continuity lp-spaces
asked Jan 3 at 8:04
ijk2438ijk2438
182
$endgroup$
add a comment |
$begingroup$
Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.
Do I need to use the definition of continuity to solve this question? I don't know how to start this question.
real-analysis analysis continuity lp-spaces
asked Jan 3 at 8:04
ijk2438ijk2438
182
$endgroup$
add a comment |
$begingroup$
Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.
Do I need to use the definition of continuity to solve this question? I don't know how to start this question.
real-analysis analysis continuity lp-spaces
asked Jan 3 at 8:04
ijk2438ijk2438
182
$endgroup$
Show that the function $$phi(r) = ||f||_{L^r(0,1)}$$ for $rin [1,2]$ is a continuous function of $r$.
Do I need to use the definition of continuity to solve this question? I don't know how to start this question.
real-analysis analysis continuity lp-spaces
real-analysis analysis continuity lp-spaces
asked Jan 3 at 8:04
ijk2438ijk2438
182
asked Jan 3 at 8:04
ijk2438ijk2438
182
asked Jan 3 at 8:04
ijk2438ijk2438
182
asked Jan 3 at 8:04
ijk2438ijk2438
182
asked Jan 3 at 8:04
ijk2438ijk2438
182
182
add a comment |
add a comment |
1 Answer
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$begingroup$
We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
$$
0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
$$ and $r mapsto |f(x)|^r$ is continuous.
Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
$$
|f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
$$ whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
$$
psi:frac{1}{p}mapsto log |f|_{p},
$$ is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.
answered Jan 3 at 8:39
SongSong
8,311625
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add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
$$
0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
$$ and $r mapsto |f(x)|^r$ is continuous.
Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
$$
|f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
$$ whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
$$
psi:frac{1}{p}mapsto log |f|_{p},
$$ is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.
answered Jan 3 at 8:39
SongSong
8,311625
$endgroup$
add a comment |
$begingroup$
We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
$$
0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
$$ and $r mapsto |f(x)|^r$ is continuous.
Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
$$
|f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
$$ whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
$$
psi:frac{1}{p}mapsto log |f|_{p},
$$ is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.
answered Jan 3 at 8:39
SongSong
8,311625
$endgroup$
add a comment |
$begingroup$
We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
$$
0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
$$ and $r mapsto |f(x)|^r$ is continuous.
Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
$$
|f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
$$ whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
$$
psi:frac{1}{p}mapsto log |f|_{p},
$$ is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.
answered Jan 3 at 8:39
SongSong
8,311625
$endgroup$
We begin by assuming that the end point cases $phi(1)$ and $phi(2)$ are both finite. This guarantees $phi(r)<infty$ for all $1le rle 2$. Let $Phi(r) = phi(r)^r$. And note that $$F(x,y) = x^{frac{1}{y}}=e^{frac{ log x}{y}} $$ is jointly continuous on $xin [0,infty), yin (0,infty)$. Since $phi(r) = F(Phi(r), r)$, it suffices to show $Phi(r)$ is continuous. Note that it is sufficient to prove $Phi(r_n)to Phi(r)$ whenever $r_n to r$. This follows easily from dominated convergence theorem by noticing that
$$
0le |f(x)|^r = |f(x)|^r 1_{{|f(x)|<1}}+|f(x)|^r 1_{{|f(x)|ge 1}}le |f(x)| 1_{{|f(x)|<1}}+|f(x)|^2 1_{{|f(x)|ge 1}}in L^1(0,1).
$$ and $r mapsto |f(x)|^r$ is continuous.
Note: However, we can say more about $phi(r)$ actually. By Holder's inequality, we can show that
$$
|f|_{{p_t}}le |f|_{{p_0}}^{1-t}|f|_{{p_1}}^t
$$ whenever $frac{1}{p_t}=frac{1-t}{p_0}+frac{t}{p_1}$, $p_0, p_1 in [1,infty]$. This shows that
$$
psi:frac{1}{p}mapsto log |f|_{p},
$$ is convex, and thus is continuous on$ {psi(x) <infty}$. Therefore, it follows that $phi(r)=e^{psi(frac{1}{r})}$ is also continuous.
answered Jan 3 at 8:39
SongSong
8,311625
edited Jan 3 at 9:14
answered Jan 3 at 8:39
SongSong
8,311625
answered Jan 3 at 8:39
SongSong
8,311625
answered Jan 3 at 8:39
SongSong
8,311625
8,311625
add a comment |
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