Let $f, g:\mathbb{R} \rightarrow \mathbb{R}$ be two continuous functions. If $f$ is convex then prove that:
$$f \left( \int_0^1 g(x) \, {\rm d}x \right) \leq \int_0^1 f\left(g(x)\right)\, {\rm d}x$$
Solution
We invoke a partition of $[0,1]$ , say $\Delta_n = \left\{ 0, \frac{1}{n}, \frac{2}{n}, \cdots, \frac{n}{n}=1 \right\}$. Since $f$ is convex it follows from Jensen's inequality that:
$$f\left ( \sum_{i=1}^{n}\frac{1}{n}g\left ( \xi_i \right ) \right )\leq \sum_{i=1}^{n}\frac{1}{n}f\left ( g(\xi_i) \right )$$
The limit procession preserves the order of the inequalities. Hence taking limits at the last equation we have that:
$$\lim_{n \rightarrow +\infty}f\left ( \sum_{i=1}^{n}\frac{1}{n}g\left ( \xi_i \right ) \right )\leq \lim_{n\rightarrow +\infty}\sum_{i=1}^{n}\frac{1}{n}f\left ( g(\xi_i) \right ) \implies f\left ( \int_{0}^{1}g(x)\, {\rm d}x \right ) = \int_{0}^{1}f\left ( g(x) \right )\, {\rm d}x$$
where we made use of the Riemann integrability in the last steps.
$$f \left( \int_0^1 g(x) \, {\rm d}x \right) \leq \int_0^1 f\left(g(x)\right)\, {\rm d}x$$
Solution
We invoke a partition of $[0,1]$ , say $\Delta_n = \left\{ 0, \frac{1}{n}, \frac{2}{n}, \cdots, \frac{n}{n}=1 \right\}$. Since $f$ is convex it follows from Jensen's inequality that:
$$f\left ( \sum_{i=1}^{n}\frac{1}{n}g\left ( \xi_i \right ) \right )\leq \sum_{i=1}^{n}\frac{1}{n}f\left ( g(\xi_i) \right )$$
The limit procession preserves the order of the inequalities. Hence taking limits at the last equation we have that:
$$\lim_{n \rightarrow +\infty}f\left ( \sum_{i=1}^{n}\frac{1}{n}g\left ( \xi_i \right ) \right )\leq \lim_{n\rightarrow +\infty}\sum_{i=1}^{n}\frac{1}{n}f\left ( g(\xi_i) \right ) \implies f\left ( \int_{0}^{1}g(x)\, {\rm d}x \right ) = \int_{0}^{1}f\left ( g(x) \right )\, {\rm d}x$$
where we made use of the Riemann integrability in the last steps.
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