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<strong>Lipschitz</strong> <strong>Constant</strong><br />

May 18, 2013<br />

1 What is <strong>Lipschitz</strong> <strong>Constant</strong><br />

First consider a single-variable function f(x) for x inside its domain D.<br />

∣ The magnitude of the slope of f(x) for 2 points (x 1 , f(x 1 )) and (x 2 , f(x 2 )) is thus<br />

f(x 1 ) − f(x 2 ) ∣∣∣<br />

∣ x 1 − x 2<br />

A non-negative real number L , which is the smallest upper bound of the slope , ( i.e. the “limiter”<br />

of the slope ) , is called the <strong>Lipschitz</strong> <strong>Constant</strong><br />

L = sup |f(x 1) − f(x 2 )|<br />

= sup<br />

df<br />

|x 1 − x 2 | ∣dx∣<br />

That is, the <strong>Lipschitz</strong> <strong>Constant</strong> limit how fast the function can change, in other words, if there is<br />

no <strong>Lipschitz</strong> <strong>Constant</strong>, the function can change extremely fast without border, that is, the function<br />

becomes discontinuous<br />

For 2-variable function<br />

For n−variable function<br />

L y = sup |f(x, y 1) − f(x, y 2 )|<br />

|y 1 − y 2 |<br />

∣ L xn = sup<br />

∂f ∣∣∣<br />

∣∂x n<br />

= sup<br />

∂f<br />

∣ ∂y ∣<br />

2 Use of <strong>Lipschitz</strong> <strong>Constant</strong> in Numerical Analysis<br />

1


3 Computation of <strong>Lipschitz</strong> <strong>Constant</strong><br />

The <strong>Lipschitz</strong> <strong>Constant</strong> can be found using definition or triangle-inequality<br />

EXAMPLE<br />

f = |y|<br />

EXAMPLE<br />

f = √ |y|<br />

∣ L = sup<br />

∂f<br />

∣∣∣ ∣ ∂y ∣ = sup ∂y ∣ ∂y ∣ = sup 1 = 1<br />

∣ L = sup<br />

∂f<br />

∣∣∣∣ ∣ ∂y ∣ = sup ∂ √ |y|<br />

∂y ∣ = sup √ 1 = ∞ /∈ R +<br />

|y|<br />

Thus there is no <strong>Lipschitz</strong> <strong>Constant</strong><br />

EXAMPLE<br />

f = x 2 y , x ∈ [−5, 2] , y ∈ [3, ∞]<br />

L = sup<br />

∂f<br />

∣ ∂y ∣ = sup ∣ ∣ x<br />

2 = 25<br />

EXAMPLE<br />

f = √ x 2 + y 2 √ √<br />

|f(x, y 1 ) − f(x, y 2 )| = | x 2 + y1 2 − x 2 + y2|<br />

2<br />

√ √<br />

√ =<br />

∣ x 2 + y1 2 − x 2 + y2<br />

2 ∣ ·<br />

x2 + y1 2 + √ x 2 + y2<br />

√ 2 x2 + y2 2 + √ x 2 + y2<br />

2<br />

=<br />

≤<br />

=<br />

|y 2 1 − y 2 2|<br />

√<br />

x2 + y 2 2 + √ x 2 + y 2 2<br />

|y 1 + y 2 |<br />

√<br />

x2 + y2 2 + √ |y 1 − y 2 |<br />

x 2 + y2<br />

2<br />

|y 1 | + |y 2 |<br />

√<br />

x2 + y2 2 + √ |y 1 − y 2 |<br />

x 2 + y2<br />

2 } {{ }<br />

≤1<br />

Thus L = 1<br />

≤ 1 · |y 1 − y 2 |<br />

−END−<br />

2

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