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“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!

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