UPSC 2023 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Riemann integral · Real Analysis · asked 10× in 13 yrs · Read the full method →

Question

Prove that the oscillation of a real-valued bounded function $f$ defined on $[a,b]$ is the supremum of the set $\{|f(x_1)-f(x_2)|:x_1,x_2\in[a,b]\}$ .

Technique

Two-sided inequality on $\sup$ — bound from above using $m\le f\le M$ , bound from below using the $\varepsilon$ -characterisation of $\sup,\inf$ .

Solution

Setup. Let $f:[a,b]\to\mathbb{R}$ be bounded. By boundedness, $M=\sup_{[a,b]}f$ and $m=\inf_{[a,b]}f$ are finite. The oscillation of $f$ on $[a,b]$ is defined as

\omega(f,[a,b])=M-m.

Define

S=\sup\bigl\{\,|f(x_1)-f(x_2)|\,:\,x_1,x_2\in[a,b]\,\bigr\}.

We must prove $S=M-m$ .

Step 1 — Upper bound: $S\le M-m$ .

For any $x_1,x_2\in[a,b]$ , $m\le f(x_i)\le M$ , so

-(M-m)\le f(x_1)-f(x_2)\le M-m\;\Longrightarrow\;|f(x_1)-f(x_2)|\le M-m.

Hence $M-m$ is an upper bound for the set $\{|f(x_1)-f(x_2)|\}$ , and by definition of supremum,

S\le M-m. \tag{1}

Step 2 — Lower bound: $S\ge M-m$ .

Fix $\varepsilon>0$ . By the definition of $\sup$ and $\inf$ :

there exists $x_1\in[a,b]$ with $f(x_1)>M-\varepsilon/2$ ;
there exists $x_2\in[a,b]$ with $f(x_2)<m+\varepsilon/2$ .

Then

f(x_1)-f(x_2)>\bigl(M-\tfrac{\varepsilon}{2}\bigr)-\bigl(m+\tfrac{\varepsilon}{2}\bigr)=(M-m)-\varepsilon,

and since this is positive (assuming $\varepsilon<M-m$ ; the trivial case $M=m$ is handled separately below),

|f(x_1)-f(x_2)|=f(x_1)-f(x_2)>(M-m)-\varepsilon.

Hence $S\ge|f(x_1)-f(x_2)|>(M-m)-\varepsilon$ . Since $\varepsilon>0$ was arbitrary,

S\ge M-m. \tag{2}

(Trivial case. If $M=m$ then $f$ is constant; both sides equal $0$ and the statement holds trivially.)

Step 3 — Combine.

From (1) and (2),

Answer

\boxed{\;\omega(f,[a,b])=M-m=\sup\bigl\{\,|f(x_1)-f(x_2)|\,:\,x_1,x_2\in[a,b]\,\bigr\}.\;}