Ytukyg

I need to show that if $|y-y_0|<text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})$ and $yneq 0$ and $y_0neq 0$ are true, then the following inequality is also true:
$$ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$$

I tried to check the information given by the inequalities $|y-y_0|<frac{|y_0|}{2}$ and $|y-y_0|<frac{epsilon|y_0|^2}{2}$. Then I arrived from those inequalities to these ones: $|frac{y}{y_0}-1|<frac{1}{2}$ and $|frac{1}{y} cdot frac{y^2}{y_0^2}-frac{1}{y_0}|<frac{epsilon}{2}$.

I multiplied the inequality $|y-y_0|<frac{epsilon|y_0|^2}{2}$ by $frac{1}{|y y_0|}$ , because $frac{1}{|yy_0|}>0$. It follows that $|frac{1}{y_0}-frac{1}{y}|<frac{epsilon|y_0|}{2|y|}$. Because $|frac{1}{y_0}-frac{1}{y}|=|frac{1}{y}-frac{1}{y_0}|$ is true for the absolute value function, we get $|frac{1}{y}-frac{1}{y_0}|<frac{epsilon|y_0|}{2|y|}$...

I am still trying to figure out, how I can prove $ leftvert frac{1}{y}-frac{1}{y_0} rightvert < epsilon.$

Any help is appreciated. Thanks in advance.

Regards,

Ahmed Hossam

Case 1:

$$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{|y_0|}{2}$$

$$frac{|y_0|}{2}lefrac{epsilon|y_0|^2}{2}$$

$$frac1{|y_0|}leepsilontag{1}$$

In this particular case:

$$|y-y_0|lt frac{|y_0|}{2}$$

$$-frac{|y_0|}{2}<y-y_0lt frac{|y_0|}{2}$$

$$y_0-frac{|y_0|}{2}<y< y_0+frac{|y_0|}{2}tag{2}$$

If $y_0>0$, (2) becomes:

$$frac{y_0}{2}<y< frac{3y_0}{2}$$

$$frac{2}{y_0}>frac1y> frac{2}{3y_0}$$

$$frac{2}{y_0}-frac{1}{y_0}>frac1y - frac{1}{y_0}> frac{2}{3y_0}-frac{1}{y_0}$$

$$frac{1}{y_0}>frac1y - frac{1}{y_0}> -frac{1}{3y_0}$$

$$frac{1}{y_0}>|frac1y - frac{1}{y_0}|tag{3}$$

Because of (1) and (3) we have:

$$epsilon>|frac1y - frac{1}{y_0}|$$

If $y_0<0$, (2) becomes:

$$y_0+frac{y_0}{2}<y< y_0-frac{y_0}{2}$$

$$frac{3y_0}{2}<y< frac{y_0}{2}$$

Notice that $y$ is squeezed between negative values so we can write:

$$frac2{3y_0}>frac1y> frac2{y_0}$$

$$frac2{3y_0}-frac{1}{y_0}>frac1y-frac1{y_0}> frac2{y_0}-frac1{y_0}$$

$$-frac{1}{3y_0}>frac1y-frac1{y_0}> frac1{y_0}$$

$$|frac1y-frac1{y_0}|< frac1{|y_0|}tag{4}$$

Because of (1) and (4) we have:

$$|frac1y-frac1{y_0}|< epsilon$$

Case 2:

$$text{min}(frac{|y_0|}{2},frac{epsilon|y_0|^2}{2})=frac{epsilon|y_0|^2}{2}$$

$$frac{|y_0|}{2}gefrac{epsilon|y_0|^2}{2})$$

$$1geepsilon|y_0|tag{5}$$

In this particular case:

$$|y-y_0|lt frac{epsilon|y_0|^2}{2}$$

$$-frac{epsilon|y_0|^2}{2}lt y-y_0lt frac{epsilon|y_0|^2}{2}$$

$$y_0-frac{epsilon|y_0|^2}{2}lt y lt y_0+frac{epsilon|y_0|^2}{2}tag{6}$$

For $y_0>0$, (6) becomes:

$$y_0-frac{epsilon y_0^2}{2}lt y lt y_0+frac{epsilon y_0^2}{2}$$

Notice that because of (5) $y$ is squeezed between positive values so we can write:

$$frac{1}{y_0-frac{epsilon y_0^2}{2}}gt frac1y gt frac1{y_0+frac{epsilon y_0^2}{2}}$$

$$frac{1}{y_0-frac{epsilon y_0^2}{2}}-frac 1{y_0} gt frac1y-frac 1{y_0} gt frac1{y_0+frac{epsilon y_0^2}{2}}-frac 1{y_0}$$

This evolves into:

$$frac{epsilon}{2-epsilon y_0}>frac1y-frac 1{y_0}>-frac{epsilon}{2+epsilon y_0}tag{7}$$

Notice that because of (5):

$$2-epsilon y_0 ge 1 implies epsilon ge frac{epsilon}{2-epsilon y_0}$$

$$2+epsilon y_0 ge 2 implies -frac{epsilon}{2+epsilon y_0}>-frac{epsilon}{2}$$

With this in mind, (7) now becomes:

$$epsilon >frac1y-frac 1{y_0}>-frac{epsilon}{2}$$

$$epsilon >|frac1y-frac 1{y_0}|$$

I'm leaving the last case ($y_0<0$) to you as an exercise.