Convexity and the Trapezoidal Approximation

Question

Trying to prove the inequality $\frac{e^b - e^a}{b-a} \leq \frac{e^a + e^b}{2}$ using convexity of $e^x$ and integration. Non-rigorously looking at the graph, the area under the curve is less than the trapezoidal area so I get $$e^b - e^a = \int_a^b e^xdx \leq \frac{1}{2}(e^a + e^b)(b-a)$$

But I want to prove for any convex and differentiable function $f:[a,b] \to \mathbb{R}$:

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

I used the mean value theorem to get

$$f(x) = f(a) + f'(\alpha(x))(x-a), \,\, a < \alpha(x)< x\\f(x) = f(b) + f'(\beta(x))(x-b), \,\, x < \beta(x)< b$$

and after combining and integrating

$$\int_a^bf(x)dx - \frac{1}{2}(f(a) + f(b))(b-a) \\= \frac{1}{2}\int_a^b f'(\alpha(x))(x-a)dx + \frac{1}{2}\int_a^b f'(\beta(x))(x-b)dx $$

(a) I'm not sure how to show the right side is $\leq 0$. I really can't say much about $\alpha$ and $\beta$

(b) Is it necessary that the derivative $f'$ be continuous?

Answer 1

This bound can be proven without any differentiation. If $f$ is convex on $[a,b]$, then it is bounded above by its secant: $$f(x)\leq \left(\frac{b-x}{b-a}\right)f(a)+\left(\frac{x-a}{b-a}\right)f(b)\text{.}$$ Consequently, $$\begin{split}\int_a^bf(x)\mathrm{d}x &\leq \int_a^b\left(\left(\frac{b-x}{b-a}\right)f(a)+\left(\frac{x-a}{b-a}\right)f(b)\right)\mathrm{d}x\text{.}\\ &=\frac{b-a}{2}f(a)+\frac{b-a}{2}f(b)\end{split}$$

Answer 2

Such inequality is just a consequence of the convexity of the exponential function, since the Hermite-Hadamard inequality ensures $$ \frac{1}{b-a}\int_{a}^{b}f(x)\,dx \leq \frac{f(a)+f(b)}{2} $$ and you can clearly see what happens by plugging in $f(x)=e^x$.

Replying to the additional questions: a convex function defined on a convex set is always continuous on the interior of its domain, but it might be non-differentiable at some point, like $|x|$ at $x=0$. On the other hand a convex function is almost-everywhere differentiable, and if $f(x)$ is convex and differentiable then $f'(x)$ is increasing. This does grant that $f'(x)$ is continuous (since by Darboux theorem derivatives have the intermediate value property, that together with injectivity implies continuity) but it does not grant that $f''(x)$ exists. On the other hand if $f(x)$ is twice differentiable and $f''(x)\geq 0$ then $f(x)$ is convex. Another useful criterion is that a midpoint-convex function $f(x)$ with a bit of extra regularity (measurability, local boundedness or continuity suffice) is convex. So the convexity of the exponential function can be seen as a straightforward consequence of $\frac{d^2}{dx^2}e^x=e^x\geq 0$ or just $\frac{e^a+e^b}{2}\geq e^{\frac{a+b}{2}}$, which is the AM-GM inequality.