Proving an integral inequality involving Lipschitz functions

Question

I am trying to prove the following inequality:

$$\int_a^b \Big( f(a)+f(b)-2f(x) \Big)dx\leq \frac{1}{2}\text{Lip}(f) (b-a)^2$$

where $f$ is a Lipschitz function and $a<b$. This inequality is numerically true for $\sin(x)$, $\cos(x)$, $\exp(x)$, polynomials and $|x|$. I wonder if it is possible to rigorously prove this inequality for all Lipschitz functions?

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However, I can prove a weaker version:

$$\int_a^b \Big( f(a)+f(b)-2f(x) \Big)dx\leq \text{Lip}(f) (b-a)^2$$

Proof:

$$|f(x)-f(y)|\leq \text{Lip}(f) |x-y| \,\forall x,y \in [a,b] \text{ and } |x-a|+|x-b|=b-a,\forall x\in [a,b]\implies$$ $$\int_a^b \Big( f(a)+f(b)-2f(x) \Big)dx \leq \int_a^b |f(a)-f(x)| + |f(b)-f(x)|\leq \text{Lip}(f) (b-a)^2$$

But this estimate is too rough to get $\frac{1} {2}\text{Lip}(f) (b-a)^2$ as an upper bound. Could you please give me some idea?

Thank you very much!

Answer 1

You know that $$f(x)\geq f(a)-\operatorname{Lip}(f)(x-a),$$ $$f(x)\geq f(b)+\operatorname{Lip}(f)(x-b).$$ So, call $$ k(x):= \max( f(a)-\operatorname{Lip}(f)(x-a), f(b)+\operatorname{Lip}(f)(x-b)), $$ and you have that $$ f(x)\geq k(x). $$ This implies $$ \int_a^b f(x)dx\geq \int_a^bk(x)dx. $$ Now you can compute the integral of $k$ explicitly, and after tedious calculations you will see that $$ \int_a^bk(x)dx\geq \frac{f(a)+f(b)}{2}(b-a)-\frac 14 \operatorname{Lip}(f)(b-a)^2. $$ Putting together the last two inequalities, you obtain the estimate that you want.

Answer 2

Thanks lorenzo Pompili for your proof! My friend gave me another proof yesterday, which is also interesting, so I leave it here:

Proof:

We can rewrite $f(b)-f(x)=\int_x^b f'(x)dx$ and $f(a)-f(x)=\int_x^a f'(x)dx$. Then we have $$ \int_a^b \Big( f(a)+f(b)-2f(x) \Big) dx = \int_a^b \Big( \int_x^b -\int_a^x \Big) f'(y) dy dx.$$ Then by changing the order of integral, we get $$\int_a^b \Big( \int_x^b -\int_a^x \Big) f'(y) dy dx=\int_a^b \Big( \int_a^y -\int_y^b \Big) f'(y) dx dy=\int_a^b f'(y) (2y-a-b)dy.$$ Since $f'(y)\leq \text{Lip}(f)$, we have $$\int_a^b f'(y) (2y-a-b)dy\leq \text{Lip}(f) \int_a^b |2y-a-b| dy.$$ After tedious calculations, we finally get $$\int_a^b |2y-a-b| dy=\frac{1}{2}(b-a)^2,$$ this conclude the proof.