Paper 1, Section II, D

Analysis I | Part IA, 2017

Let a,bRa, b \in \mathbb{R} with a<ba<b and let f:(a,b)Rf:(a, b) \rightarrow \mathbb{R}.

(a) Define what it means for ff to be continuous at y0(a,b)y_{0} \in(a, b).

ff is said to have a local minimum at c(a,b)c \in(a, b) if there is some ε>0\varepsilon>0 such that f(c)f(x)f(c) \leqslant f(x) whenever x(a,b)x \in(a, b) and xc<ε|x-c|<\varepsilon.

If ff has a local minimum at c(a,b)c \in(a, b) and ff is differentiable at cc, show that f(c)=0f^{\prime}(c)=0.

(b) ff is said to be convex if

f(λx+(1λ)y)λf(x)+(1λ)f(y)f(\lambda x+(1-\lambda) y) \leqslant \lambda f(x)+(1-\lambda) f(y)

for every x,y(a,b)x, y \in(a, b) and λ[0,1]\lambda \in[0,1]. If ff is convex, rRr \in \mathbb{R} and [y0r,y0+r](a,b)\left[y_{0}-|r|, y_{0}+|r|\right] \subset(a, b), prove that

(1+λ)f(y0)λf(y0r)f(y0+λr)(1λ)f(y0)+λf(y0+r)(1+\lambda) f\left(y_{0}\right)-\lambda f\left(y_{0}-r\right) \leqslant f\left(y_{0}+\lambda r\right) \leqslant(1-\lambda) f\left(y_{0}\right)+\lambda f\left(y_{0}+r\right)

for every λ[0,1]\lambda \in[0,1].

Deduce that if ff is convex then ff is continuous.

If ff is convex and has a local minimum at c(a,b)c \in(a, b), prove that ff has a global minimum at cc, i.e., that f(x)f(c)f(x) \geqslant f(c) for every x(a,b)x \in(a, b). [Hint: argue by contradiction.] Must ff be differentiable at cc ? Justify your answer.

Typos? Please submit corrections to this page on GitHub.