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$M$ equicontinuous and closed set of $C([a,b],mathbbR)$. There is $m geq 0$ such that $|f(a)| leq m$ $forall f$. Prove that $M$ is compact
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Problem. Let $mathcalM$ an equicontinuous and closed set of $C([a,b],mathbbR)$. Suppose that there is a $m geq 0$ such that $|f(a)| leq m$ $forall f in mathcalM$. Show that there are $f_1$ and $f_2 in mathcalM$ such that
$$int_a^bf_1(x)mathrmdx leq int_a^bf(x)mathrmdx leq int_a^bf_2(x)mathrmdx,quad forall f in mathcalM.$$
My ideia is define a function $varphi: mathcalM to mathbbR$ by $displaystyle varphi(f) = int_a^bf(x)dx$ and to conclude using de compactness of $mathcalM$. For show that $mathcalM$ is compact, I tried to use the Arzelà-Ascoli Theorem. I just to show that $mathcalM(x) = f(x) mid f in mathcalM $ is bounded and this is my problem. By hypothesis, $mathcalM(a)$ is bounded and probably, I should to use it, but I don't see how. Can someone help me?
real-analysis continuity arzela-ascoli
asked 4 hours ago
Lucas Corrêa
1,041219
add a comment |Â
up vote
0
down vote
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Problem. Let $mathcalM$ an equicontinuous and closed set of $C([a,b],mathbbR)$. Suppose that there is a $m geq 0$ such that $|f(a)| leq m$ $forall f in mathcalM$. Show that there are $f_1$ and $f_2 in mathcalM$ such that
$$int_a^bf_1(x)mathrmdx leq int_a^bf(x)mathrmdx leq int_a^bf_2(x)mathrmdx,quad forall f in mathcalM.$$
My ideia is define a function $varphi: mathcalM to mathbbR$ by $displaystyle varphi(f) = int_a^bf(x)dx$ and to conclude using de compactness of $mathcalM$. For show that $mathcalM$ is compact, I tried to use the Arzelà-Ascoli Theorem. I just to show that $mathcalM(x) = f(x) mid f in mathcalM $ is bounded and this is my problem. By hypothesis, $mathcalM(a)$ is bounded and probably, I should to use it, but I don't see how. Can someone help me?
real-analysis continuity arzela-ascoli
asked 4 hours ago
Lucas Corrêa
1,041219
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Problem. Let $mathcalM$ an equicontinuous and closed set of $C([a,b],mathbbR)$. Suppose that there is a $m geq 0$ such that $|f(a)| leq m$ $forall f in mathcalM$. Show that there are $f_1$ and $f_2 in mathcalM$ such that
$$int_a^bf_1(x)mathrmdx leq int_a^bf(x)mathrmdx leq int_a^bf_2(x)mathrmdx,quad forall f in mathcalM.$$
My ideia is define a function $varphi: mathcalM to mathbbR$ by $displaystyle varphi(f) = int_a^bf(x)dx$ and to conclude using de compactness of $mathcalM$. For show that $mathcalM$ is compact, I tried to use the Arzelà-Ascoli Theorem. I just to show that $mathcalM(x) = f(x) mid f in mathcalM $ is bounded and this is my problem. By hypothesis, $mathcalM(a)$ is bounded and probably, I should to use it, but I don't see how. Can someone help me?
real-analysis continuity arzela-ascoli
asked 4 hours ago
Lucas Corrêa
1,041219
Problem. Let $mathcalM$ an equicontinuous and closed set of $C([a,b],mathbbR)$. Suppose that there is a $m geq 0$ such that $|f(a)| leq m$ $forall f in mathcalM$. Show that there are $f_1$ and $f_2 in mathcalM$ such that
$$int_a^bf_1(x)mathrmdx leq int_a^bf(x)mathrmdx leq int_a^bf_2(x)mathrmdx,quad forall f in mathcalM.$$
My ideia is define a function $varphi: mathcalM to mathbbR$ by $displaystyle varphi(f) = int_a^bf(x)dx$ and to conclude using de compactness of $mathcalM$. For show that $mathcalM$ is compact, I tried to use the Arzelà-Ascoli Theorem. I just to show that $mathcalM(x) = f(x) mid f in mathcalM $ is bounded and this is my problem. By hypothesis, $mathcalM(a)$ is bounded and probably, I should to use it, but I don't see how. Can someone help me?
real-analysis continuity arzela-ascoli
asked 4 hours ago
Lucas Corrêa
1,041219
asked 4 hours ago
Lucas Corrêa
1,041219
asked 4 hours ago
Lucas Corrêa
1,041219
asked 4 hours ago
Lucas Corrêa
1,041219
1,041219
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
1
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Your idea is fine. Since $mathcal M$ is equicontinuous, there is a $delta>0$ such that, for each $x,yin[a,b]$ and each $finmathcal M$, $|x-y|<deltaimpliesbigl|f(x)-f(y)bigr|<1$. Now, take $Ninmathbb N$ such that $a+Ndeltageqslant b$. Then, for each $xin[a,b]$ and each $finmathcal M$, $f(x)leqslant m+N$, because $aleqslant xleqslant b$ and therefore $x=a+kdelta+x'$ for some $k<N$ and some $x'in[0,delta]$, which implies thatbeginalignbigl|f(x)bigr|&=bigl|f(a+kdelta+x')bigr|\&leqslantbigl|f(a)bigr|+bigl|f(a+delta)-f(a)bigr|+cdots+bigl|f(a+kdelta+x')-f(a+kdelta)bigr|\&leqslantbigl|f(a)bigr|+k+1\&leqslantbigl|f(a)bigr|+N.endalignSo, the hypothesis of the Arzelà-Ascoli hold and therefore $mathcal M$ is indeed compact.
answered 4 hours ago
José Carlos Santos
111k1695171
I apologize, but I don't see why $a + Ndelta geq b$ implies $|f(x)| leq m + N$
– Lucas Corrêa
2 hours ago
1
@LucasCorrêa I've edited my answer. I hope that everything is clear and correct now.
– José Carlos Santos
2 hours ago
Is perfect! I was trying $|f(x)| leq |f(a)| + |f(x) - f(a)|$.
– Lucas Corrêa
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your idea is fine. Since $mathcal M$ is equicontinuous, there is a $delta>0$ such that, for each $x,yin[a,b]$ and each $finmathcal M$, $|x-y|<deltaimpliesbigl|f(x)-f(y)bigr|<1$. Now, take $Ninmathbb N$ such that $a+Ndeltageqslant b$. Then, for each $xin[a,b]$ and each $finmathcal M$, $f(x)leqslant m+N$, because $aleqslant xleqslant b$ and therefore $x=a+kdelta+x'$ for some $k<N$ and some $x'in[0,delta]$, which implies thatbeginalignbigl|f(x)bigr|&=bigl|f(a+kdelta+x')bigr|\&leqslantbigl|f(a)bigr|+bigl|f(a+delta)-f(a)bigr|+cdots+bigl|f(a+kdelta+x')-f(a+kdelta)bigr|\&leqslantbigl|f(a)bigr|+k+1\&leqslantbigl|f(a)bigr|+N.endalignSo, the hypothesis of the Arzelà-Ascoli hold and therefore $mathcal M$ is indeed compact.
answered 4 hours ago
José Carlos Santos
111k1695171
I apologize, but I don't see why $a + Ndelta geq b$ implies $|f(x)| leq m + N$
– Lucas Corrêa
2 hours ago
1
@LucasCorrêa I've edited my answer. I hope that everything is clear and correct now.
– José Carlos Santos
2 hours ago
Is perfect! I was trying $|f(x)| leq |f(a)| + |f(x) - f(a)|$.
– Lucas Corrêa
2 hours ago
add a comment |Â
up vote
1
down vote
accepted
Your idea is fine. Since $mathcal M$ is equicontinuous, there is a $delta>0$ such that, for each $x,yin[a,b]$ and each $finmathcal M$, $|x-y|<deltaimpliesbigl|f(x)-f(y)bigr|<1$. Now, take $Ninmathbb N$ such that $a+Ndeltageqslant b$. Then, for each $xin[a,b]$ and each $finmathcal M$, $f(x)leqslant m+N$, because $aleqslant xleqslant b$ and therefore $x=a+kdelta+x'$ for some $k<N$ and some $x'in[0,delta]$, which implies thatbeginalignbigl|f(x)bigr|&=bigl|f(a+kdelta+x')bigr|\&leqslantbigl|f(a)bigr|+bigl|f(a+delta)-f(a)bigr|+cdots+bigl|f(a+kdelta+x')-f(a+kdelta)bigr|\&leqslantbigl|f(a)bigr|+k+1\&leqslantbigl|f(a)bigr|+N.endalignSo, the hypothesis of the Arzelà-Ascoli hold and therefore $mathcal M$ is indeed compact.
answered 4 hours ago
José Carlos Santos
111k1695171
I apologize, but I don't see why $a + Ndelta geq b$ implies $|f(x)| leq m + N$
– Lucas Corrêa
2 hours ago
1
@LucasCorrêa I've edited my answer. I hope that everything is clear and correct now.
– José Carlos Santos
2 hours ago
Is perfect! I was trying $|f(x)| leq |f(a)| + |f(x) - f(a)|$.
– Lucas Corrêa
2 hours ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your idea is fine. Since $mathcal M$ is equicontinuous, there is a $delta>0$ such that, for each $x,yin[a,b]$ and each $finmathcal M$, $|x-y|<deltaimpliesbigl|f(x)-f(y)bigr|<1$. Now, take $Ninmathbb N$ such that $a+Ndeltageqslant b$. Then, for each $xin[a,b]$ and each $finmathcal M$, $f(x)leqslant m+N$, because $aleqslant xleqslant b$ and therefore $x=a+kdelta+x'$ for some $k<N$ and some $x'in[0,delta]$, which implies thatbeginalignbigl|f(x)bigr|&=bigl|f(a+kdelta+x')bigr|\&leqslantbigl|f(a)bigr|+bigl|f(a+delta)-f(a)bigr|+cdots+bigl|f(a+kdelta+x')-f(a+kdelta)bigr|\&leqslantbigl|f(a)bigr|+k+1\&leqslantbigl|f(a)bigr|+N.endalignSo, the hypothesis of the Arzelà-Ascoli hold and therefore $mathcal M$ is indeed compact.
answered 4 hours ago
José Carlos Santos
111k1695171
Your idea is fine. Since $mathcal M$ is equicontinuous, there is a $delta>0$ such that, for each $x,yin[a,b]$ and each $finmathcal M$, $|x-y|<deltaimpliesbigl|f(x)-f(y)bigr|<1$. Now, take $Ninmathbb N$ such that $a+Ndeltageqslant b$. Then, for each $xin[a,b]$ and each $finmathcal M$, $f(x)leqslant m+N$, because $aleqslant xleqslant b$ and therefore $x=a+kdelta+x'$ for some $k<N$ and some $x'in[0,delta]$, which implies thatbeginalignbigl|f(x)bigr|&=bigl|f(a+kdelta+x')bigr|\&leqslantbigl|f(a)bigr|+bigl|f(a+delta)-f(a)bigr|+cdots+bigl|f(a+kdelta+x')-f(a+kdelta)bigr|\&leqslantbigl|f(a)bigr|+k+1\&leqslantbigl|f(a)bigr|+N.endalignSo, the hypothesis of the Arzelà-Ascoli hold and therefore $mathcal M$ is indeed compact.
answered 4 hours ago
José Carlos Santos
111k1695171
edited 2 hours ago
answered 4 hours ago
José Carlos Santos
111k1695171
answered 4 hours ago
José Carlos Santos
111k1695171
answered 4 hours ago
José Carlos Santos
111k1695171
111k1695171
I apologize, but I don't see why $a + Ndelta geq b$ implies $|f(x)| leq m + N$
– Lucas Corrêa
2 hours ago
1
@LucasCorrêa I've edited my answer. I hope that everything is clear and correct now.
– José Carlos Santos
2 hours ago
Is perfect! I was trying $|f(x)| leq |f(a)| + |f(x) - f(a)|$.
– Lucas Corrêa
2 hours ago
add a comment |Â
I apologize, but I don't see why $a + Ndelta geq b$ implies $|f(x)| leq m + N$
– Lucas Corrêa
2 hours ago
1
@LucasCorrêa I've edited my answer. I hope that everything is clear and correct now.
– José Carlos Santos
2 hours ago
Is perfect! I was trying $|f(x)| leq |f(a)| + |f(x) - f(a)|$.
– Lucas Corrêa
2 hours ago
I apologize, but I don't see why $a + Ndelta geq b$ implies $|f(x)| leq m + N$
– Lucas Corrêa
2 hours ago
I apologize, but I don't see why $a + Ndelta geq b$ implies $|f(x)| leq m + N$
– Lucas Corrêa
2 hours ago
1
1
@LucasCorrêa I've edited my answer. I hope that everything is clear and correct now.
– José Carlos Santos
2 hours ago
@LucasCorrêa I've edited my answer. I hope that everything is clear and correct now.
– José Carlos Santos
2 hours ago
Is perfect! I was trying $|f(x)| leq |f(a)| + |f(x) - f(a)|$.
– Lucas Corrêa
2 hours ago
Is perfect! I was trying $|f(x)| leq |f(a)| + |f(x) - f(a)|$.
– Lucas Corrêa
2 hours ago
add a comment |Â
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