How to prove the continuity of a function $mathbb{R}to L^1(mathbb{R}^n)$.
$begingroup$
Lets say I have a function $varphi:mathbb{R}to L^1(mathbb{R}^n)$ which maps $t$ into $varphi(t)=[xmapsto f(t,x)]$. Proving continuity of such function amounts to proving that for all $varepsilon>0$, there exists $delta>0$ such that
$$|t_1-t_2|<delta implies int_{mathbb{R}^n}|f(t_1,x)-f(t_2,x)|:mathrm{d}x<varepsilon.$$
If it were
$$|t_1-t_2|<delta implies left|int_{mathbb{R}^n}[f(t_1,x)-f(t_2,x)]:mathrm{d}xright|<varepsilon,$$
I could use the theorem on the continuity of a parametric integral. However what I want to prove is stronger than that. How should I approach it?
real-analysis measure-theory lebesgue-integral
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
$endgroup$
add a comment |
$begingroup$
Lets say I have a function $varphi:mathbb{R}to L^1(mathbb{R}^n)$ which maps $t$ into $varphi(t)=[xmapsto f(t,x)]$. Proving continuity of such function amounts to proving that for all $varepsilon>0$, there exists $delta>0$ such that
$$|t_1-t_2|<delta implies int_{mathbb{R}^n}|f(t_1,x)-f(t_2,x)|:mathrm{d}x<varepsilon.$$
If it were
$$|t_1-t_2|<delta implies left|int_{mathbb{R}^n}[f(t_1,x)-f(t_2,x)]:mathrm{d}xright|<varepsilon,$$
I could use the theorem on the continuity of a parametric integral. However what I want to prove is stronger than that. How should I approach it?
real-analysis measure-theory lebesgue-integral
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
$endgroup$
$begingroup$
You could find a function $g_deltain L^1(mathbb R^n)$ such that $|f(t_1,x)-f(t_2,x)|leq g_delta(x)$, where $deltageq |t_1-t_2|$ and then show that $int |g_delta|$ is bounded by some $M>0$. If $g_delta$ is decreasing in $delta$, it should be pretty straight forward. Also note that what you want to do there is proving uniform continuity, which is stronger than just continuity.
$endgroup$
– Pink Panther
Jan 5 at 10:29
add a comment |
$begingroup$
Lets say I have a function $varphi:mathbb{R}to L^1(mathbb{R}^n)$ which maps $t$ into $varphi(t)=[xmapsto f(t,x)]$. Proving continuity of such function amounts to proving that for all $varepsilon>0$, there exists $delta>0$ such that
$$|t_1-t_2|<delta implies int_{mathbb{R}^n}|f(t_1,x)-f(t_2,x)|:mathrm{d}x<varepsilon.$$
If it were
$$|t_1-t_2|<delta implies left|int_{mathbb{R}^n}[f(t_1,x)-f(t_2,x)]:mathrm{d}xright|<varepsilon,$$
I could use the theorem on the continuity of a parametric integral. However what I want to prove is stronger than that. How should I approach it?
real-analysis measure-theory lebesgue-integral
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
$endgroup$
Lets say I have a function $varphi:mathbb{R}to L^1(mathbb{R}^n)$ which maps $t$ into $varphi(t)=[xmapsto f(t,x)]$. Proving continuity of such function amounts to proving that for all $varepsilon>0$, there exists $delta>0$ such that
$$|t_1-t_2|<delta implies int_{mathbb{R}^n}|f(t_1,x)-f(t_2,x)|:mathrm{d}x<varepsilon.$$
If it were
$$|t_1-t_2|<delta implies left|int_{mathbb{R}^n}[f(t_1,x)-f(t_2,x)]:mathrm{d}xright|<varepsilon,$$
I could use the theorem on the continuity of a parametric integral. However what I want to prove is stronger than that. How should I approach it?
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
asked Jan 5 at 10:12
Gabriel RibeiroGabriel Ribeiro
1,454523
1,454523
$begingroup$
You could find a function $g_deltain L^1(mathbb R^n)$ such that $|f(t_1,x)-f(t_2,x)|leq g_delta(x)$, where $deltageq |t_1-t_2|$ and then show that $int |g_delta|$ is bounded by some $M>0$. If $g_delta$ is decreasing in $delta$, it should be pretty straight forward. Also note that what you want to do there is proving uniform continuity, which is stronger than just continuity.
$endgroup$
– Pink Panther
Jan 5 at 10:29
add a comment |
$begingroup$
You could find a function $g_deltain L^1(mathbb R^n)$ such that $|f(t_1,x)-f(t_2,x)|leq g_delta(x)$, where $deltageq |t_1-t_2|$ and then show that $int |g_delta|$ is bounded by some $M>0$. If $g_delta$ is decreasing in $delta$, it should be pretty straight forward. Also note that what you want to do there is proving uniform continuity, which is stronger than just continuity.
$endgroup$
– Pink Panther
Jan 5 at 10:29
$begingroup$
You could find a function $g_deltain L^1(mathbb R^n)$ such that $|f(t_1,x)-f(t_2,x)|leq g_delta(x)$, where $deltageq |t_1-t_2|$ and then show that $int |g_delta|$ is bounded by some $M>0$. If $g_delta$ is decreasing in $delta$, it should be pretty straight forward. Also note that what you want to do there is proving uniform continuity, which is stronger than just continuity.
$endgroup$
– Pink Panther
Jan 5 at 10:29
$begingroup$
You could find a function $g_deltain L^1(mathbb R^n)$ such that $|f(t_1,x)-f(t_2,x)|leq g_delta(x)$, where $deltageq |t_1-t_2|$ and then show that $int |g_delta|$ is bounded by some $M>0$. If $g_delta$ is decreasing in $delta$, it should be pretty straight forward. Also note that what you want to do there is proving uniform continuity, which is stronger than just continuity.
$endgroup$
– Pink Panther
Jan 5 at 10:29
add a comment |
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$begingroup$
You could find a function $g_deltain L^1(mathbb R^n)$ such that $|f(t_1,x)-f(t_2,x)|leq g_delta(x)$, where $deltageq |t_1-t_2|$ and then show that $int |g_delta|$ is bounded by some $M>0$. If $g_delta$ is decreasing in $delta$, it should be pretty straight forward. Also note that what you want to do there is proving uniform continuity, which is stronger than just continuity.
$endgroup$
– Pink Panther
Jan 5 at 10:29