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# 2.2 Proof of the Cauchy–Riemann Converse Theorem

The proof of the Cauchy–Riemann Converse Theorem is rather involved and may require more than one reading.

We will need two results from real analysis. The first result is known as the Mean Value Theorem.

## Theorem 6 Mean Value Theorem

Let be a real function that is continuous on the closed interval  and differentiable on the open interval . Then there is a number such that

Equation label: (equation 6)

To appreciate why this theorem is true, imagine pushing the chord between and in Figure 13 parallel to itself until it becomes a tangent to the graph of at a point , where lies somewhere between and . Clearly, the gradient of the original chord must be equal to the gradient of the tangent, so

Multiplication by gives . Notice that this equation is also true if .

Figure 13 Graph of the real function

The second result that we will need is a Linear Approximation Theorem, which asserts that if is a real-valued function of two real variables and , then for near , the value of can be approximated by the value of the linear function defined by

Now, the graph of is a plane passing through the point on the graph of (Figure 14). Moreover, the partial - and -derivatives of coincide with the partial - and -derivatives of at . This means that both have the same gradient in the - and -directions, so you can think of the plane as the tangent plane to the graph of at .

Figure 14 Tangent plane to the graph of at the point

The accuracy with which this tangent plane approximates the graph of depends on the smoothness of the graph of . If the graph exhibits the kind of kink shown in Figure 12, then the approximation is not as good as for a function with continuous partial derivatives.

## Theorem 7 Linear Approximation Theorem (double-struck cap r squared to double-struck cap r)

Let be a real-valued function of two real variables, defined on a region in containing . If the partial - and -derivatives of  exist on and are continuous at , then there is an ‘error function’ such that

where as .

Since is the distance from to , the theorem asserts that the error function tends to zero ‘faster’ than this distance. Theorem 7 is the real-valued function analogue of Theorem 2.

## Proof

We have to show that the function  defined by

satisfies

Since the partial derivatives exist on , they must be defined on some disc centred at . Let us begin by finding an expression for on this disc. If we apply the Mean Value Theorem to the real functions (where is kept constant) and , then we obtain

where is between and , and

where is between and (see Figure 15). Hence

Figure 15 Vertical and horizontal line segments from to

Substituting this expression for into the definition of , we obtain

Dividing both sides by , and noting that

(because both and do not exceed ), we see that

Figure 15 illustrates that as tends to , so do and . So, by the continuity of the partial - and -derivatives at , the two terms on the right of the inequality above must both tend to 0 as tends to . It follows that tends to 0 as tends to .

We are now in a position to prove the Cauchy–Riemann Converse Theorem.

## Theorem 5 Cauchy–Riemann Converse Theorem (revisited)

Let be defined on a region containing . If the partial derivatives , , ,

• exist at for each

• are continuous at

• satisfy the Cauchy–Riemann equations at ,

then is differentiable at and

## Proof

We need to show that the limit of the difference quotient for at exists and has the value indicated in the theorem. In order to calculate the difference quotient for at , we find an expression for . Since and fulfil the conditions of Theorem 7, it follows that

where and are the error functions associated with and , respectively.

Collecting together terms, we see that

Since and satisfy the Cauchy–Riemann equations, both expressions in the large brackets must be equal, so

Dividing by gives

The limit of this difference quotient exists, and has the required value

provided that we can show that the expression involving the error functions and tends to 0 as tends to . To this end, notice that is equal to and so, by the Triangle Inequality,

By Theorem 7, both expressions on the right tend to 0 as tends to . Consequently, the expression on the left must also tend to 0, and the theorem follows.