I cannot prove that if f(x) is convex on [a,b] then f(a+b2)≤1b−a∫abf(x)dx≤f(a)+f(b)2

The inequality f(a+b2)≤1b−a∫abf(x)dx is a special case of Jensens

Since f is convex ob [a,b], one has: ∀x∈[a,b], f(x)≤f(b)−f(a)b−a(x−a)+f(a) Therefore, one gets: ∫abf(x)dx≤f(b)−f(a)b−a(b−a)22+f(a)(b−a) Finally, one has: 1b−a∫abf(x)dx≤f(b)+f(a)2

The inequality f(a+b2)≤1b−a∫abf(x)dx is a special case of Jensens

Since f is convex ob [a,b], one has: ∀x∈[a,b], f(x)≤f(b)−f(a)b−a(x−a)+f(a) Therefore, one gets: ∫abf(x)dx≤f(b)−f(a)b−a(b−a)22+f(a)(b−a) Finally, one has: 1b−a∫abf(x)dx≤f(b)+f(a)2

Get the solution to "I cannot prove that if f(x) is convex..."

Algotssleeddynf

Answered question

2022-01-05

I cannot prove that if

f (x)

is convex on

[a, b]

then

f (\frac{a + b}{2}) \leq \frac{1}{b - a} \int_{a}^{b} f (x) d x \leq \frac{f (a) + f (b)}{2}

Answer & Explanation

otoplilp1

Beginner2022-01-06Added 41 answers

The inequality

f (\frac{a + b}{2}) \leq \frac{1}{b - a} \int_{a}^{b} f (x) d x

is a special case of Jensens

sukljama2

Beginner2022-01-07Added 32 answers

For convenience, we can take

b = 1, a = - 1

For the first inequality, use

f (0) \leq \frac{f (x) + f (- x)}{2}

and integrate both sides from

x = - 1

to 1
For the second, use

f (x) = f (\frac{1 - x}{2} (- 1) + \frac{1 + x}{2} (1)) \leq \frac{1 - x}{2} f (- 1) + \frac{1 + x}{2} f (1)

and integrate both sides from

x = - 1

to 1.

star233

Skilled2022-01-11Added 403 answers

Since f is convex ob [a,b], one has:
$\forall x \in [a, b], f (x) \leq \frac{f (b) - f (a)}{b - a} (x - a) + f (a)$
Therefore, one gets:
$\int_{a}^{b} f (x) d x \leq \frac{f (b) - f (a)}{b - a} \frac{(b - a)^{2}}{2} + f (a) (b - a)$
Finally, one has:
$\frac{1}{b - a} \int_{a}^{b} f (x) d x \leq \frac{f (b) + f (a)}{2}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?