Question

Let $f:R\to R$ be a function such that $f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}$ for all x, y, and $f\left(0\right)=3$ and $f^{\prime }\left(0\right)=3$. Then

• A. $f\left(x\right)/x$ is continuous on R
• B. f(x) is continuous on R
• C. f(x) is bounded on R
• D. none of these

Answer 1

Answer: (B)

f(x) is continuous on R

$\underset{h\to 0}{lim}f(x+h)=\underset{h\to 0}{lim}f\left(\frac{2x+2h}{2}\right)$

$=\underset{h\to 0}{lim}\frac{1}{2}\left[f\right(2x)+f(2h\left)\right]$

$=\frac{1}{2}f\left(2x\right)+\frac{1}{2}\underset{h\to 0}{lim}f\left(2h\right)=\frac{1}{2}f\left(2x\right)+\frac{1}{2}f\left(0\right)$

(f is differentiable at 0 so continuous also)
Putting $y=0$ in the given equation, we have
$f\left(x\right)=f\left(\frac{2x}{2}\right)=\frac{f\left(2x\right)+f\left(0\right)}{2}$

Hence $\underset{h\to 0}{lim}f(x+h)=f\left(x\right)$

Since $f\left(x\right)/x$ is not defined at $x=0,f\left(x\right)/x$ is not continuous on R. Clearly $f\left(x\right)$ need not be bounded on R.

e.g. $f\left(x\right)=x$ satisfies the given equation but is not bounded on R.