I cannot prove that if f(x) is convex on [a,b]

Algotssleeddynf

Algotssleeddynf

Answered question

2022-01-05

I cannot prove that if f(x) is convex on [a,b] then
f(a+b2)1baabf(x)dxf(a)+f(b)2

Answer & Explanation

otoplilp1

otoplilp1

Beginner2022-01-06Added 41 answers

The inequality
f(a+b2)1baabf(x)dx
is a special case of Jensens
sukljama2

sukljama2

Beginner2022-01-07Added 32 answers

For convenience, we can take b=1, a=1
For the first inequality, use
f(0)f(x)+f(x)2
and integrate both sides from x=1 to 1
For the second, use
f(x)=f(1x2(1)+1+x2(1))1x2f(1)+1+x2f(1)
and integrate both sides from x=1 to 1.
star233

star233

Skilled2022-01-11Added 403 answers

Since f is convex ob [a,b], one has:
x[a,b], f(x)f(b)f(a)ba(xa)+f(a)
Therefore, one gets:
abf(x)dxf(b)f(a)ba(ba)22+f(a)(ba)
Finally, one has:
1baabf(x)dxf(b)+f(a)2

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