← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q4a — Step-by-Step Solution

20 marks · Section A

Real number system as ordered field with LUB property · Real Analysis · asked 2× in 14 yrs · Read the full method →

Question

Suppose $\mathbb R$ be the set of all real numbers and $f:\mathbb R\to\mathbb R$ is a function such that the following equations hold for all $x,y\in\mathbb R$:

• (i) $f(x+y)=f(x)+f(y)$
• (ii) $f(xy)=f(x)f(y)$

Show that $\forall x\in\mathbb R$ either $f(x)=0$, or, $f(x)=x$.

Technique

Ring-homomorphism structure: (ii) at $x=y=1$ splits into zero/identity cases; additivity fixes $\mathbb Q$; (ii) gives squares $\to$ non-negatives $\Rightarrow$ monotonicity; density of $\mathbb Q$ + monotonicity $\Rightarrow$ $f=\mathrm{id}$.

Solution

We must show the only functions satisfying both Cauchy’s additive equation (i) and the multiplicative equation (ii) are the zero map and the identity map (and that one of these holds globally, not pointwise-mixed).

Step 1 — $f(1)=0$ or $f(1)=1$

Put $x=y=1$ in (ii): $f(1)=f(1)^2$, so $f(1)\big(f(1)-1\big)=0$, giving

Case $f(1)=0$. For any $x$, (ii) with $y=1$ gives $f(x)=f(x\cdot1)=f(x)f(1)=0$. So $f\equiv0$ — the zero map. From here assume $f(1)=1$.

Step 2 — $f$ fixes the integers, then the rationals

From (i), $f(0)=f(0)+f(0)\Rightarrow f(0)=0$, and $f(-x)=-f(x)$. By induction, additivity gives $f(n)=nf(1)=n$ for all $n\in\mathbb Z$.

For a rational $q=p/n$ ($p\in\mathbb Z,n\in\mathbb N$): additivity gives $f(n\cdot q)=nf(q)$, and $n\cdot q=p$, so $nf(q)=f(p)=p$, hence

So $f$ is the identity on $\mathbb Q$.

Step 3 — $f$ is monotone (order-preserving) via squares

The key extra ingredient beyond Cauchy: $f$ maps squares to squares, hence non-negatives to non-negatives. For $t\ge0$, write $t=s^2$; then by (ii)

Thus $x\ge0\Rightarrow f(x)\ge0$. Now if $x\le y$ then $y-x\ge0$, so $f(y)-f(x)=f(y-x)\ge0$ (using additivity and $f(-x)=-f(x)$). Therefore

Step 4 — Monotone + identity on $\mathbb Q$ forces $f=\mathrm{id}$

Let $x\in\mathbb R$ be arbitrary. Choose rationals $q_1\le x\le q_2$. By monotonicity and Step 2,

Since rationals are dense, we may let $q_1\uparrow x$ and $q_2\downarrow x$; squeezing gives

So in the case $f(1)=1$, $f$ is the identity.

Conclusion

Either $f\equiv0$ (case $f(1)=0$) or $f(x)=x$ for all $x$ (case $f(1)=1$). In both cases, for every $x$, $f(x)=0$ or $f(x)=x$:

Answer