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1

$begingroup$

Consider a sequence of continuous function $f_n:[a,b]to mathbbR$. Suppose there exist constants $gamma>1$ and $beta>0$ independent of $n,p$ such that
$$left(int_a^b|f_n(x)|^pgammadxright)^frac1gammaleq pcdotbeta^frac1pleft(int_a^b|f_n(x)|^pdxright)^fracp-1p$$ for any $p>1$ and $nin mathbbN$. Show that there exists a constant $C$ depending on $gamma$ and $beta$, but independent of $n$, such that $$max_xin[a,b]|f_n(x)|leq Cleft(sqrtint_a^b f_n(x)^2dx+1right)$$ for any $n in mathbbN$.

A hint has been given: Suppose $f(x)geq 0$ is continuous on $[a,b]$, we have $$lim_pto +infty left(int_a^bf(x)^pdxright)^frac1p=max_[a,b]f(x)$$

real-analysis limits sequence-of-function

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asked Apr 8 at 1:23

VladimirVladimir

62

$endgroup$

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1

$begingroup$

Consider a sequence of continuous function $f_n:[a,b]to mathbbR$. Suppose there exist constants $gamma>1$ and $beta>0$ independent of $n,p$ such that
$$left(int_a^b|f_n(x)|^pgammadxright)^frac1gammaleq pcdotbeta^frac1pleft(int_a^b|f_n(x)|^pdxright)^fracp-1p$$ for any $p>1$ and $nin mathbbN$. Show that there exists a constant $C$ depending on $gamma$ and $beta$, but independent of $n$, such that $$max_xin[a,b]|f_n(x)|leq Cleft(sqrtint_a^b f_n(x)^2dx+1right)$$ for any $n in mathbbN$.

A hint has been given: Suppose $f(x)geq 0$ is continuous on $[a,b]$, we have $$lim_pto +infty left(int_a^bf(x)^pdxright)^frac1p=max_[a,b]f(x)$$

real-analysis limits sequence-of-function

share|cite|improve this question

asked Apr 8 at 1:23

VladimirVladimir

62

$endgroup$

add a comment | 

$begingroup$

Consider a sequence of continuous function $f_n:[a,b]to mathbbR$. Suppose there exist constants $gamma>1$ and $beta>0$ independent of $n,p$ such that
$$left(int_a^b|f_n(x)|^pgammadxright)^frac1gammaleq pcdotbeta^frac1pleft(int_a^b|f_n(x)|^pdxright)^fracp-1p$$ for any $p>1$ and $nin mathbbN$. Show that there exists a constant $C$ depending on $gamma$ and $beta$, but independent of $n$, such that $$max_xin[a,b]|f_n(x)|leq Cleft(sqrtint_a^b f_n(x)^2dx+1right)$$ for any $n in mathbbN$.

A hint has been given: Suppose $f(x)geq 0$ is continuous on $[a,b]$, we have $$lim_pto +infty left(int_a^bf(x)^pdxright)^frac1p=max_[a,b]f(x)$$

real-analysis limits sequence-of-function

share|cite|improve this question

asked Apr 8 at 1:23

VladimirVladimir

62

$endgroup$

share|cite|improve this question

asked Apr 8 at 1:23

VladimirVladimir

62

share|cite|improve this question

asked Apr 8 at 1:23

VladimirVladimir

62

asked Apr 8 at 1:23

VladimirVladimir

62

asked Apr 8 at 1:23

VladimirVladimir

62