Exercise 0.7. Show that (9) (for proper f(-)) is equivalent to the seemingly weaker condition f(xx) ≤ (1-X)f(x) + f(x₁) where Io, 11 € dom(f) and 0≤x≤ 1. (10)

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Exercise 0.7. Show that (9) (for proper f()) is equivalent to the seemingly weaker condition
where xo, 1 € dom(f) and 0≤x≤ 1. (10)
f(xx) ≤ (1-X)f(x) + f(x1)
The definition of a convex function f(.) without assuming differentiability (or even being
finite-valued) is given as follows: Suppose f(): R → R. Then f() is convex provided for all
To, 1 and 0 ≤ ≤ 1, we have
f(xx) ≤ (1-X) f(xo) + Af(x₁)
where zx = (1-X)xo + A₁.
Actually, though, with this definition, if f(x) = -x for some x then some of the anomalies
alluded to above occur, so let us restrict to only proper functions. We note that since f(.)
may take on the value +∞o, the inequality in (9) must be interpreted with the conventions of
extended arithmetic (see item 6). The following exercise says, however, that only at points
where f() is less than infinity need to be considered.
Transcribed Image Text:Exercise 0.7. Show that (9) (for proper f()) is equivalent to the seemingly weaker condition where xo, 1 € dom(f) and 0≤x≤ 1. (10) f(xx) ≤ (1-X)f(x) + f(x1) The definition of a convex function f(.) without assuming differentiability (or even being finite-valued) is given as follows: Suppose f(): R → R. Then f() is convex provided for all To, 1 and 0 ≤ ≤ 1, we have f(xx) ≤ (1-X) f(xo) + Af(x₁) where zx = (1-X)xo + A₁. Actually, though, with this definition, if f(x) = -x for some x then some of the anomalies alluded to above occur, so let us restrict to only proper functions. We note that since f(.) may take on the value +∞o, the inequality in (9) must be interpreted with the conventions of extended arithmetic (see item 6). The following exercise says, however, that only at points where f() is less than infinity need to be considered.
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