Question

(a)

Suppose first that x and y are both non-negative. Then f(x) = x, f(y) = y, and it is straightforward that, for all α ∈ [0, 1],

f(αx + (1 − α)y) = αx + (1 − α)y = αf(x) + (1 − α)f(y).

Thus, the condition for convexity is satisfied in this case. The case when x and y are both negative is also easy. The only case remaining to consider is when one is positive and the other negative. Suppose that x < 0 and y > 0, and let αin[0, 1]. Now,

f(αx+(1−α)y)=|αx+(1−α)y| ≤ |αx|+|(1−α)y|
= α|x|+(1−α)|y|=αf(x)+(1−α)f(y).

The inequality in the above equation is called the triangle inequality, and is known to hold for all real (and complex) numbers.

Answer 1

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