Continuity of the Derivative

We show now that the differential approximation

f[x + δx] - f[x] = f^′[x] · δx + ε · δx

forces the derivative function f ' [x] to be continuous,

x_1≈x_2 ⇒ f^′ [x_1] ≈f^′[x_2]

Let x_1≈x_2,  but x_1≠x_2.  Use the differential approximation with x = x_1 and δx = x_2 - x_1 and also with x = x_2and δx = x_1 - x_2, geometrically, looking at the tangent approximation from both endpoints.

f[x_2] - f[x_1] = f^′ [x_1] · (x_2 - x_1) + ε_1 · (x_2 - x_1)

f[x_1] - f[x_2] = f^′ [x_2] · (x_1 - x_2) + ε_2 · (x_1 - x_2)

Adding, we obtain

0 = ((f^′ [x_1] - f^′ [x_2]) + (ε_1 - ε_2)) · (x_2 - x_1)

Dividing by the nonzero term (x_2 - x_1) and adding f ' [x_2] to both sides, we obtain,

f^′ [x_2] = f^′ [x_1] + (ε_1 - ε_2) or f^′[x_2] ≈f^′[x_1], since the difference between two small errors is small.

This fact can be used to prove:

Theorem: The Inverse Function Theorem

If f^′[x_0] ≠0 then f[x] has an inverse function in a small neighborhood of x_0, that is, if y≈y_0 = f[x_0], then there is a unique x≈x_0so that y = f[x].

We saw above that the differential approximation makes a microscopic view of the graph look linear.  If y≈y_1 the linear equation dy = m · dx with m = f ' [x_1] can be inverted to find a first approximation to the inverse,

y - y_0 = m · (x_1 - x_0)

x_1 = x_0 + 1/m (y - y_0)

[Graphics:../HTMLFiles/Lect1_122.gif]

We test to see if f[x_1] = y.  If not, examine the graph microscopically at (x_1, y_1) = (x_1, f[x_1]).  Since the graph appears the same as it's tangent to within ε and since m = f ' [x_1] ≈f ' [x_2], the local coordinates at (x_1, y_1) look like a line of slope m:

[Graphics:../HTMLFiles/Lect1_129.gif]

Solving for the x-value which gives output y, we get

y - y_1 = m · (x_2 - x_1)

x_2 = x_1 + 1/m (y - f[x_1])

Continue in this way generating a sequence of approximations, x_1 = x_0 + 1/m (y - y_0), x_ (n + 1) = G[x_n], where the recursion function G[ξ] = x + 1/m (y - f[ξ]).  The distance between successive approximations is

| x_2 - x_1 | = | G[x_1] - G[x_0] | ≤ | G ' [ξ_1] | · | x_1 - x_0 |

| x_3 - x_2 | = | G[x_2] - G[x_1] | ≤ | G ' [ξ_2] | · | x_2 - x_1 | ≤ | G ' [ξ_2] | · | G ' [ξ_1] | · | x_1 - x_0 |

by the Mean Value Theorem for Derivatives.  Notice that G ' [ξ] = 1 - f ' [ξ]/m≈0, for ξ≈x_0, so | G ' [ξ] | <1/2 in particular, and

| x_2 - x_1 | = | G[x_1] - G[x_0] | ≤1/2 · | x_1 - x_0 |

| x_3 - x_2 | = | G[x_2] - G[x_1] | ≤1/2 · | x_2 - x_1 | ≤1/2^2 · | x_1 - x_0 |

:

| x_ (n + 1) - x_n | ≤1/2^n · | x_1 - x_0 |

| x_ (n + 1) - x_0 | ≤ | x_ (n + 1) - x_n | +| x_n - x_ (n - 1) | +⋯ + | x_1 - x_0 | ≤ (1 + 1/2 + ⋯ + 1/2^n) · | x_1 - x_0 |

A geometric series estimate shows that the series converges, x_n x≈x_0 and f[x] = y.

To complete this proof we need to show that G[ξ] is a contraction on some nonzero interval.  The function G[ξ] must map the interval into itself and have derivative less than 1/2 on the interval.  The precise definiton of the derivative matters because the result is false if f^′[x] is defined by a pointwise limit.  The function f[x] = x + x^2Sin[π/x] with f[0] = 0 has pointwise derivative 1 at zero, but is not increasing in any neighborhood of zero.  (If you "move the microscope an infinitesimal amount" when looking at y = x^2Sin[π/x] + x, the graph will look nonlinear.)

Created by Mathematica  (September 22, 2004)