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Proof of the Integral Form of Jensen's Inequality 📂Lemmas

Proof of the Integral Form of Jensen's Inequality

Theorem

Given a convex function ϕ:[a,b]R \phi : [a,b] \to \mathbb{R} and f:[0,1][a,b]f: [0,1] \to [a,b], if ϕf\phi \circ f is integrable over [0,1][0,1], then ϕ(01f(x)dx)01(ϕf)(x)dx \phi \left( \int_{0}^{1} f(x) dx \right) \le \int_{0}^{1} (\phi \circ f ) (x) dx

Explanation

Of course, given the conditions, the integration interval can also be changed through substitution, etc. Unlike finite form, which generalizes the number of terms using definitions, integration form becomes an inequality that the function traverses over the integration symbol.

Proof

By the Mean Value Theorem for Integrals, for some constant cc, we can set 01f(x)dx=c(a,b)\displaystyle \int_{0}^{1} f(x) dx = c \in (a,b). According to the definition of cc, ϕ(01f(x)dx)=ϕ(c)+s(01f(x)dxc) \phi \left( \int_{0}^{1} f(x) dx \right) = \phi (c) + s \left( \int_{0}^{1} f(x) dx - c \right) Since the above equation holds for all sRs \in \mathbb{R}, s=supx[a,c)ϕ(c)ϕ(x)cx\displaystyle s = \sup_{x \in [a,c) } {{\phi (c) - \phi (x)} \over {c -x}} is fine to be defined. For [c,y][a,b][c, y] \subset [a,b] satisfying yy, since f(x)f(c)xcf(y)f(x)yx\displaystyle {{f(x) - f(c)} \over {x - c}} \le {{f(y) - f(x) } \over {y - x}}, sϕ(y)ϕ(c)ycs \le {{\phi (y) - \phi (c)} \over {y- c}} Reorganizing the found equation, ϕ(c)+s(yc)ϕ(y) \phi (c) + s (y - c) \le \phi (y) Meanwhile, in the case of [y,c][a,b][y, c] \subset [a,b], sϕ(c)ϕ(y)cy s \ge {{\phi (c) - \phi (y)} \over {c - y}} That is, for all y[a,b]y \in [a,b], ϕ(c)+s(yc)ϕ(y)\phi (c) + s (y - c) \le \phi (y) holds and [a,b][a,b] is the range of ff, so it can be put as y=f(x)y = f(x) to get the following. ϕ(c)+s(f(x)c)ϕ(f(x)) \phi (c) + s ( f(x) - c) \le \phi ( f(x) ) Taking 01\displaystyle \int_{0}^{1} to both sides, 01{ϕ(c)+s(f(x)c)}dx01(ϕ(f(x))dx \int_{0}^{1} \left\{ \phi (c) + s ( f(x) - c) \right\} dx \le \int_{0}^{1} (\phi (f (x) ) dx and organizing yields ϕ(01f(x)dx)+s01f(x)dxsc01(ϕf)(x)dx \phi \left( \int_{0}^{1} f(x) dx \right) + s \int_{0}^{1} f(x) dx - s c \le \int_{0}^{1} (\phi \circ f) (x) dx Lastly, since 01f(x)dx=c\displaystyle \int_{0}^{1} f(x) dx = c was true, it is s01f(x)dxsc=0\displaystyle s \int_{0}^{1} f(x) dx - s c = 0, and we obtain the following inequality we wanted. ϕ(01f(x)dx)01(ϕf)(x)dx \phi \left( \int_{0}^{1} f(x) dx \right) \le \int_{0}^{1} (\phi \circ f) (x) dx

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