What's the reason behind $e^{tx} \leq \frac{b-x}{b-a}e^{ta}+\frac{x-a}{b-a}e^{tb}$?

Question

$e^{tx}$ is a convex function of $x$

Then, $\forall x\in [a,b]$

$e^{tx} \leq \frac{b-x}{b-a}e^{ta}+\frac{x-a}{b-a}e^{tb}$ holds.

Which is not same as convex definition $f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2)$.

An explanation of my inequality would be appreciated. Thank you very much in advance.

Answer 1

You are using the letter $t$ twice for two different things. Instead use $$f(\lambda x_1 + (1-\lambda)x_2) \le \lambda f(x_1) + (1-\lambda) f(x_2).$$

Let $x_1 = a$, $x_2 = b$, and $\lambda = \dfrac{b-x}{b-a}$.

Answer 2

Note that for a convex function

$$f(\lambda x + (1-\lambda) y)\le \lambda f(x) + (1-\lambda) f(y) $$

and in this case

• $x=at$
• $y=bt$
• $\lambda= \frac{b-x}{b-a}$