Theorem or Justification for the function with a given finite integral over an interval with the lowwest...












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In a paper I'm writing, I'm trying to find a formal theorem or other concise way of stating that in the space of continuous functions whose definite integral over the over the interval $,0 leq tleq t_f$ is equal to a given constant, positive value,
$int_0^{t_f},mathcal{f}(t)=C$ , the function $mathcal{f}(t)$ with the lowest maximum value is a constant. The principle seems self evident but for my purposes I would like to have a way to show this definitively. Because I'm currently going off of the self evidence of the thing, I could very well be wrong, so I would also like to have a definitive answer before proceeding in the paper but I've been struggling to find any information on this problem, or to demonstrate it myself.










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  • $begingroup$
    This is false if you allow discontinuous functions.
    $endgroup$
    – Randall
    Jan 11 at 18:45






  • 1




    $begingroup$
    You are correct! Thank you very much. I've edited the question to allow only for continuous functions.
    $endgroup$
    – MargaretL
    Jan 11 at 18:49
















0












$begingroup$


In a paper I'm writing, I'm trying to find a formal theorem or other concise way of stating that in the space of continuous functions whose definite integral over the over the interval $,0 leq tleq t_f$ is equal to a given constant, positive value,
$int_0^{t_f},mathcal{f}(t)=C$ , the function $mathcal{f}(t)$ with the lowest maximum value is a constant. The principle seems self evident but for my purposes I would like to have a way to show this definitively. Because I'm currently going off of the self evidence of the thing, I could very well be wrong, so I would also like to have a definitive answer before proceeding in the paper but I've been struggling to find any information on this problem, or to demonstrate it myself.










share|cite|improve this question











$endgroup$












  • $begingroup$
    This is false if you allow discontinuous functions.
    $endgroup$
    – Randall
    Jan 11 at 18:45






  • 1




    $begingroup$
    You are correct! Thank you very much. I've edited the question to allow only for continuous functions.
    $endgroup$
    – MargaretL
    Jan 11 at 18:49














0












0








0





$begingroup$


In a paper I'm writing, I'm trying to find a formal theorem or other concise way of stating that in the space of continuous functions whose definite integral over the over the interval $,0 leq tleq t_f$ is equal to a given constant, positive value,
$int_0^{t_f},mathcal{f}(t)=C$ , the function $mathcal{f}(t)$ with the lowest maximum value is a constant. The principle seems self evident but for my purposes I would like to have a way to show this definitively. Because I'm currently going off of the self evidence of the thing, I could very well be wrong, so I would also like to have a definitive answer before proceeding in the paper but I've been struggling to find any information on this problem, or to demonstrate it myself.










share|cite|improve this question











$endgroup$




In a paper I'm writing, I'm trying to find a formal theorem or other concise way of stating that in the space of continuous functions whose definite integral over the over the interval $,0 leq tleq t_f$ is equal to a given constant, positive value,
$int_0^{t_f},mathcal{f}(t)=C$ , the function $mathcal{f}(t)$ with the lowest maximum value is a constant. The principle seems self evident but for my purposes I would like to have a way to show this definitively. Because I'm currently going off of the self evidence of the thing, I could very well be wrong, so I would also like to have a definitive answer before proceeding in the paper but I've been struggling to find any information on this problem, or to demonstrate it myself.







calculus






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share|cite|improve this question













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share|cite|improve this question








edited Jan 11 at 18:48







MargaretL

















asked Jan 11 at 18:44









MargaretLMargaretL

11




11












  • $begingroup$
    This is false if you allow discontinuous functions.
    $endgroup$
    – Randall
    Jan 11 at 18:45






  • 1




    $begingroup$
    You are correct! Thank you very much. I've edited the question to allow only for continuous functions.
    $endgroup$
    – MargaretL
    Jan 11 at 18:49


















  • $begingroup$
    This is false if you allow discontinuous functions.
    $endgroup$
    – Randall
    Jan 11 at 18:45






  • 1




    $begingroup$
    You are correct! Thank you very much. I've edited the question to allow only for continuous functions.
    $endgroup$
    – MargaretL
    Jan 11 at 18:49
















$begingroup$
This is false if you allow discontinuous functions.
$endgroup$
– Randall
Jan 11 at 18:45




$begingroup$
This is false if you allow discontinuous functions.
$endgroup$
– Randall
Jan 11 at 18:45




1




1




$begingroup$
You are correct! Thank you very much. I've edited the question to allow only for continuous functions.
$endgroup$
– MargaretL
Jan 11 at 18:49




$begingroup$
You are correct! Thank you very much. I've edited the question to allow only for continuous functions.
$endgroup$
– MargaretL
Jan 11 at 18:49










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