← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q2c — Step-by-Step Solution

15 marks · Section A

Continuity of Functions on R; Epsilon-Delta · Real Analysis · Read the full method →

Question

Show that if a function $f$ defined on an open interval $(a,b)$ of $\mathbb R$ is convex, then $f$ is continuous. Show, by example, if the condition of open interval is dropped, then the convex function need not be continuous.

Technique

Three-point (increasing chord-slope) characterization of convexity ⇒ local two-sided slope bounds ⇒ local Lipschitz ⇒ continuity; counterexample via an endpoint jump on a closed interval.

Solution

Recall $f:(a,b)\to\mathbb R$ is convex if for all $x,y\in(a,b)$ and $\lambda\in[0,1]$,

An equivalent and convenient form is the three-point slope inequality: for $x<y<z$ in $(a,b)$,

(The slopes of chords increase from left to right.)

Step 1 — Fix an interior point and trap it strictly inside

Let $x_0\in(a,b)$. Since $(a,b)$ is open, choose points

We show $f$ is continuous at $x_0$ by bounding the difference quotients on both sides between two fixed slopes.

Step 2 — Bound the slopes near $x_0$

Take any $x$ with $p<x<x_0$ and any $y$ with $x_0<y<q$. Apply the slope monotonicity (S) repeatedly. Using the points $p<x<x_0$:

Using the points $x_0<y<q$ together with $p$:

The key conclusion is that for $x,y$ in a fixed neighborhood $[p,q]$ of $x_0$, the slopes of chords through $x_0$ are bounded between the fixed constants $K_L$ and $K_R$ (both finite, since $f(p),f(x_0),f(q)$ are finite numbers). Concretely there is a constant $M=\max(|K_L|,|K_R|)<\infty$ with

Step 3 — Lipschitz bound forces continuity

From the slope bound,

Hence as $x\to x_0$, $f(x)\to f(x_0)$. So $f$ is continuous at $x_0$. Since $x_0\in(a,b)$ was arbitrary, $f$ is continuous on $(a,b)$.

In fact this shows more: $f$ is locally Lipschitz, hence continuous, on the open interval. $\blacksquare$

Answer