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Aug 16, 2019 · Check the Hessian matrix of the function. If the matrix is: Positive-definite then your function is strictly convex. Positive semi-definite then your function is convex. A matrix is positive definite when all the eigenvalues are positive and semi-definite if all the eigenvalues are positive or zero-valued.
Answer. Prove that each of the following functions is convex on the given domain: f (x)=e^ {b x}, x \in \mathbb {R}, where b is a constant. f (x)=x^ {k}, x \in [0, \infty) and k \geq 1 is a constant. f (x)=-\ln (1-x), x \in (-\infty, 1). f (x)=-\ln \left (\frac {e^ {x}} {1+e^ {x}}\right), x \in \mathbb {R}.
A twice differentiable function of several variables is convex on a convex set if and only if its Hessian matrix of second partial derivatives is positive semidefinite on the interior of the convex set. For a convex function the sublevel sets and with are convex sets.
In fact, a ne functions are the only functions that are both convex and concave. Some quadratic functions: f(x) = xTQx+ cTx+ d. { Convex if and only if Q 0. { Strictly convex if and only if Q˜0. { Concave if and only if Q 0; strictly concave if and only if Q˚0. { The proofs are easy if we use the second order characterization of convexity (com-
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If the functions \(f\) and \(g\) are convex downward (upward), then any linear combination \(af + bg\) where \(a\), \(b\) are positive real numbers is also convex downward (upward). If the function \(u = g\left( x \right)\) is convex downward, and the function \(y = f\left( u \right)\) is convex downward and non-decreasing, then the composite ...
Recall a function f: [a;b] !R is said to be convex if f( x+ (1 )y) f(x) + (1 )f(y) for any x;y2[a;b] and any 2[0;1]. Moreover, fis called strictly convex if it is convex and f( x+ (1 )y) = f(x) + (1 )f(y) implies x= yor 2f0;1g. Now we prove the following result. Proposition 1. Let f: [a;b] !R be a di erentiable convex function. Then f Xn i=1 p ...
f is a convex function. EXAMPLES. Theorem 2 implies that both f(x) = x2 and f(x) = ex are convex because their second derivatives are the positive valued functions 2 (the constant function) and ex respectively. Similarly, f(x) = 1=x is convex on the open half-line de ned by x > 0 because f00(x) = 2=x3 is positive for x > 0.
