设f: [-1,1]→R为偶函数,在 [0,1]上单调递增,又设g是[-1,1]上的凸函数

hbghlyj

f(x) is an even function on the interval [-1, 1] and monotonically increasing on [0,1].
g(x) is a convex function on the interval [-1, 1], meaning for any x, y∈[-1, 1] and t∈(0, 1), the inequality g(tx + (1-t)y) ≤ tg(x) + (1-t)g(y) holds.
Prove:
2 * ∫_-1¹ f(x)g(x) dx ≥ ∫_-1¹ f(x) dx * ∫_-1¹ g(x) dx

hbghlyj

Since $f(x)$ is an even function, $f(-x) = f(x)$. This allows us to write:
\[
\int_{-1}^{1} f(x)g(x) \, dx = 2 \int_{0}^{1} f(x)g(x) \, dx.
\]
Let $G(x) = g(x) + g(-x)$. For a convex function $g(x)$, $G(x)$ is increasing on $[0, 1]$. This follows because for $|p| \leq |q| \leq 1$, convexity implies:
\[
g(p) + g(-p) \leq g(q) + g(-q).
\]
Using the symmetry of $f(x)$ and the definition of $G(x)$, we rewrite the integral:
\[
\int_{-1}^{1} f(x)g(x) \, dx = \int_{0}^{1} f(x) \big( g(x) + g(-x) \big) \, dx = \int_{0}^{1} f(x)G(x) \, dx.
\]
Similarly, the integrals of $f(x)$ and $g(x)$ over $[-1, 1]$ become:
\[
\int_{-1}^{1} f(x) \, dx = 2 \int_{0}^{1} f(x) \, dx, \quad \int_{-1}^{1} g(x) \, dx = \int_{0}^{1} G(x) \, dx.
\]
Since $f(x)$ and $G(x)$ are increasing on $[0, 1]$, Chebyshev inequality implies:
\[
\int_{0}^{1} f(x)G(x) \, dx \geq \left( \int_{0}^{1} f(x) \, dx \right) \left( \int_{0}^{1} G(x) \, dx \right).
\]
Substituting back, we have:
\[
2 \int_{-1}^{1} f(x)g(x) \, dx \geq \left( \int_{-1}^{1} f(x) \, dx \right) \left( \int_{-1}^{1} g(x) \, dx \right).
\]