Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$
Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.
$(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and
$$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$
$(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$
Answer:
$(i)$
Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.
Thus,
$ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.
Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have
$M-epsilon leq |f(x)|$.
Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then
$ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.
From $(1)$ and $(2)$, we have
$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.
I need confirmation of my work.
Also help me with part $(ii)$.
real-analysis integration
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
add a comment |
Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.
$(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and
$$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$
$(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$
Answer:
$(i)$
Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.
Thus,
$ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.
Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have
$M-epsilon leq |f(x)|$.
Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then
$ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.
From $(1)$ and $(2)$, we have
$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.
I need confirmation of my work.
Also help me with part $(ii)$.
real-analysis integration
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
add a comment |
Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.
$(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and
$$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$
$(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$
Answer:
$(i)$
Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.
Thus,
$ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.
Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have
$M-epsilon leq |f(x)|$.
Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then
$ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.
From $(1)$ and $(2)$, we have
$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.
I need confirmation of my work.
Also help me with part $(ii)$.
real-analysis integration
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
Suppose $f$ is a continuous function on $[a,b]$ and let $M=sup_{ a leq x leq b} |f(x)|$.
$(i)$ If $M>0$ and $p$ be any positive constant, show that for every $ epsilon>0$ there are constant $ c<d$ so that $ [c,d] subset [a,b]$ and
$$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a) .$$
$(ii)$ Prove that $ lim_{p to infty} left(int_a^b |f(x)|^pdx right)^{frac{1}{p}}.$
Answer:
$(i)$
Since $ M=sup_{ a leq x leq b} |f(x)|$, we have $ |f(x)| leq M$.
Thus,
$ int_a^b |f(x)|^pdx leq M^p int_a^b dx=M^p(b-a), ............(1)$.
Also since $ M=sup_{ a leq x leq b} |f(x)|$, by definition of supremum, we have
$M-epsilon leq |f(x)|$.
Also suppose that $ M-epsilon$ value is attained in the subset $[c,d]$, then
$ int_a^b |f(x)|^p dx geq int_c^d |f(x)|^pdx geq (M- epsilon)^p int_c^d dx=(M-epsilon)^p (d-c), ............(2)$.
From $(1)$ and $(2)$, we have
$ (M-epsilon)^p (d-c) leq int_a^b |f(x)|^pdx leq M^p(b-a)$.
I need confirmation of my work.
Also help me with part $(ii)$.
real-analysis integration
real-analysis integration
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
asked Nov 22 '18 at 22:16
arifamatharifamath
1176
1176
add a comment |
add a comment |
1 Answer
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It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
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1 Answer
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1 Answer
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It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
It is not true that $M-epsilon$ must be attained at some point. For example if $fequiv M$ then $M-epsilon$ is not attained. The correct argument is as follows: there exists $x$ such that $|f(x)| >M-epsilon$. By continuity there exists an interval $[c,d]$ around $x$ on which $|f| >M-epsilon$. Now proceed as you have done. This gives i). For ii) simply raise all sides of the inequality in i) to power $frac 1 p$ and take the limit. Use the fact that $t^{1/p} to 1$ as $p to infty$ for any $t>0$. (Use this for $t=d-c$ and $t=b-a$). By sandwich theorem you get ii).
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 22 '18 at 23:24
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
51.8k32055
add a comment |
add a comment |
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