"Upper" bound

Let $f:\mathbb{R}^+ \rightarrow \mathbb{R}$ be a function satisfying

$\left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a}$

for all positive real numbers $a$ and $b$ . Prove that

$\left | f(1) - f(x) \right | \leq \left |\ln x \right | \quad \text{for all} \;\; x>0$

Solution

For starters, let us assume that $x>1$ . Dividing the interval $(1, x)$ into $n$ subintervals each of length $h$ so that $h=\frac{x-1}{n}$ . Thus,

$\begin{align*} \left | f(1) - f(x) \right | &= \left | \sum_{k=0}^{n-1} f \left ( 1 + h(k+1) \right ) - f \left ( 1+kh \right ) \right | \\ &\leq \sum_{k=0}^{n-1} \left | f \left ( 1 + h(k+1) \right ) - f \left ( 1+kh \right ) \right | \\ &=\sum_{k=0}^{n-1} \left | f\left ( \left ( 1+kh \right ) + h\right ) - f \left ( 1+kh \right ) \right | \end{align*}$

The inequality $\left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a}$ implies that

$\left | f\left ( \left ( 1+kh \right ) + h\right ) - f \left ( 1+kh \right ) \right | \leq \frac{h}{1+kh}$

Hence,

$\left | f(1) - f(x) \right | \leq \sum_{k=0}^{n-1} \frac{h}{1+kh}$

The limit $\displaystyle \sum_{k=0}^{n-1} \frac{h}{1+kh}$ exists and equals to $\ln x$ . Hence , the inequality is proved for $x>1$ .

Now, assume that $x<1$ . Dividing the interval $(x, 1)$ into $n$ subintervals each of length $h$ so that $h=\frac{1-x}{n}$ . Thus,

$\begin{align*} \left | f(1) - f(x) \right | &= \left | \sum_{k=0}^{n-1} f \left ( 1 - h(k+1) \right ) - f \left ( 1 - kh \right ) \right | \\ &\leq \sum_{k=0}^{n-1} \left | f \left ( 1 - h(k+1) \right ) - f \left ( 1 - kh \right ) \right | \\ &=\sum_{k=0}^{n-1} \left | f\left ( \left ( 1 - kh \right ) - h\right ) - f \left ( 1 - kh \right ) \right | \end{align*}$

The inequality $\left | f\left ( a + b \right ) - f(a)\right | \leq \frac{b}{a}$ implies that

$\left | f\left ( \left ( 1 - kh \right ) - h\right ) - f \left ( 1 - kh \right ) \right | \leq \frac{h}{1-kh}$

Hence,

$\left | f(1) - f(x) \right | \leq \sum_{k=0}^{n-1} \frac{h}{1+kh}$

The limit $\displaystyle \sum_{k=0}^{n-1} \frac{h}{1-kh}$ exists and equals to $-\ln x$ . Hence , the inequality is also proved for $x<1$ . This completes the proof!

“Upper” bound

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