Inequality with indefinite integrals and extreme points











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f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?










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  • Use integration by parts. Perhaps that can help with estimation.
    – Anurag A
    Nov 18 at 19:17















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f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?










share|cite|improve this question






















  • Use integration by parts. Perhaps that can help with estimation.
    – Anurag A
    Nov 18 at 19:17













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up vote
0
down vote

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f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?










share|cite|improve this question













f:[0,1]->R is continuous and nonconstant function. F is the indefinite integral of f such that F(0)=F(1)=0. m and M are the minimum and the maximum values of f.Now we define g:[0,1]->R , g(x)=xf(x) and the indefinite integral of g,G, such that G(0)=0. How can you prove that $|G(1)|<=-mM/(2(M-m))$ ?







calculus functions boundary-value-problem






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asked Nov 18 at 19:13









Andrei Gabor

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  • Use integration by parts. Perhaps that can help with estimation.
    – Anurag A
    Nov 18 at 19:17


















  • Use integration by parts. Perhaps that can help with estimation.
    – Anurag A
    Nov 18 at 19:17
















Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17




Use integration by parts. Perhaps that can help with estimation.
– Anurag A
Nov 18 at 19:17















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