is square root convex

is square root convex sqrt x is indeed sure concave in its domain i e 0 infty but arctan x in concave only for x 0 the inteval where the last inequality you wrote holds but for x

In mathematics a real valued function is called convex if the line segment between any two distinct points on the graph of the function lies above or on the graph between the two points Equivalently a function is convex if its epigraph the set of points on or above the graph of the function is a convex set In simple terms a convex function graph is shaped like a cup or a straight line like a linear fun Root on the positive axis is a reciprocal on the negative axis while the convex conjugate of a reciprocal on the positive axis is a square root on the negative axis

is square root convex

is square root convex
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I am trying to prove that if f mathbb R n mapsto mathbb R is a strictly convex function then g x sqrt f x will not be necessarily convex I have started with To see that the converse statements are not true observe that f x x is convex but not strictly convex and f x x4 is strictly convex but not strongly convex why A ne functions f x

Consider the counterexample f x x 3 2 It is clearly convex and yet its square root x 3 4 is not convex An even simpler counterexample is f x x as long as you allow for the line The square root function is concave so you could just use a series of cutting planes linearisation of square root at sum limits i a i x i text sth le theta j with enough

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The vector 2 norm square kxk2 Euclidean norm square and Frobenius norm square kXk 2 F for example are strictly convex functions of their respective argument A function f in one variable de ned on an interval I R is convex if f 00 x 0 for all x 2I and concave if f 00 x 0 for all x 2I The graph of convex and concave function have the following shapes

The square function is convex and non monotone for arguments of unknown sign It can take the affine expression x as an argument the result square x is convex The arithmetic operator is affine and increasing so the composition Let f R 0 be a convex function If f is twice differentiable then f2 2ff 2 f 2 2ff which is 0 since f f 0 But how can I prove that f2 is convex without

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is square root convex - To see that the converse statements are not true observe that f x x is convex but not strictly convex and f x x4 is strictly convex but not strongly convex why A ne functions f x