We begin a proof of Riemann mapping theorem, which will not be concluded. But this is the now standard proof, given in many textbooks, so finish it off by reading. In the proof, a proper simply connected, connected open subset U of C is given. If there is a biholomorphic mapping from U to D₁(0), then (since we know Aut(D₁(0) very well now) we can require that the mapping take a selected point p in U to 0 in D₁(0), and we can also ask that the derivative of this mapping at p be real. The Schwarz Lemma also suggests that we make this derivative as large as possible, so that the image will be all of the disc (this is just a suggestion!). Many proofs of the RMT create a sequence of functions {f_n}, and the desired biholomorphic mapping is the limit of the sequence. We should try to see when such sequences must have a limit.

Problem Given a "collection" of functions F, when is it true that every sequence in that collection has a subsequence which converges uniformly?

A good collection of examples is important. Our function space for examples was C[0,1], the continuous functions on the unit interval. For simplicity, we will consider real-valued continuous functions.

Example 1 The family of functions is the sequence f_n(x)=n. Certainly this sequence does not converge uniformly. From this we learn that we'd better require that our family of functions is bounded (sometimes called uniformly bounded): that is, there is B>0 so that for all f's in F, we know that |f(x)|<=B. The graphs of f(x) are between y=-B and y=B.

Example 2 The family of functions is the sequence f_n(x)=sin(nx). I claim that this sequence does not converge uniformly, and, indeed, that it has no subsequence which converges uniformly. Well, just look at x=0 and what happens near 0. If there is a function g which is a limit of a subsequence of the f_n's, then I certainly know that g(0)=0 since all of the f_n's have value 0 at 0. Notice that for some very large n, |f_n(x)-g(x)|<1/2 for x's near 0. Well, since g(0)=0, I bet that we can select an interval of x's near 0 ("some [0,D] for D>0") so |g(x)|<1/2. Then for large enough n and for those x's, |f_n(x)|<=|f_n(x)-g(x)|+|g(x)|<1/2+1/2=1. But, golly, f_n has period 2Pi/n. For large n's, this period certainly fits inside [0,D]. And the values of sine on a period are [-1,1], so there is no uniformly convergent subsequence of this sequence of functions. The sequence of functions wiggles too much.

In general, if a sequence of functions converges uniformly to a limit function, the limit function's wiggle must be close to the wiggle of almost all of the sequence. Of course, it would be helpful if the word or idea of wiggle were made more precise, and, in a real variables courses, I am sure that you will all learn about the oscillation of a function. For this course, we just need the following definition.

Definition F is equicontinuous at x₀ if any e>0, there exists d>0, such that for any f in F, if |x-x₀|<d, then |f(x)-f(x₀)|<e.

A true fact (what's a false fact?) is that a collection of continuous functions defined on a compact metric space which is equicontinuous at every point is actually uniformly equicontinuous (e doesn't depend on x₀). This follows from standard compactness arguments.

Now we can state a famous theorem which says the necessary conditions mentioned are actually sufficient. To me, this result is not obvious. It needed consideration of many more examples than the few we have considered.

Theorem (Arzela-Ascoli) If F is a subset of C[0,1] which is bounded and (uniformly) equicontinuous, then every sequence in F has a uniformly convergent subsequences.

You should try to prove this yourself, even if you've heard or read or have sat through (!) a proof. Just look at a {x_n} which is dense in [0,1], and create a subsequence of any sequence in F (diagonally selected!) converge everywhere uniformly.

How can we guarantee or "force" that a collection of functions is equicontinuous at (even) one point? One simple condition follows:

Suppose F is a subset of C[0,1], and suppose that all of the functons in F are differentiable. IF there is a constant K so that |f´(x)|<=K then F is equicontinuous. This follows from the Mean Value Theorem. MVT says f(b)-f(a)=f´(c)(b-a). Now if |b-a|<e/K, |f(b)-f(a)| will certainly be less than e. So d can be e/K, the same for all of the f's in F.

This is not the only way to verify equicontinuity, of course. One can ask for uniform Lipschitz conditions: there's q>0 and r>0 so that |f(b)-f(a)|<=r|b-a|^q (think of sqrt(x) and companions on [0,1]: certainly not Lipschitz (a q=1 condition won't be satisfied near 0 but a q=1/2 will be) and certainly not differentiable. There was some discussion of this earlier in the diary.

Notice that we can put together intervals to get convergence on all of R. For example, here is a result depending A-A for closed bounded intervals:

Suppose F is a collection of continuous functions on R and that we know F is bounded on any compact interval in R, and also that F is uniformly equicontinuous on any compact interval in R. Then every sequence in f has a uniformly convergent subsequence.

Why is this true? Take any sequence of functions in F. We apply A-A on [-1,1] and get a uniformly convergence subsequence. Then consider this subsequence on [-2,2] and get a subsequence of that subsequence, and then ... [-n,n] ... and choose the final subsequence to be a diagonal subsequence of this sequence of subsequences. The result is a sequence which converges uniformly on any compact subset of R (not necessarily on R itself, though). So we use the sigma-compactness of R. Any subset of C is also sigma-compact, so the same approach will be successful there.

What happens when we consider holomorphic functions? For example, suppose we have a function f holomorphic in a disc, with |f´(z)|<=K. If z_a and z_b are in the disc, then f(z_b)-f(z_a)=INT_pathf´(z) dz. We can take the path to be, for example, a straight line segment from z_a to z_b. But ML then gives |f(z_b)-f(z_a|<=K|z_b-z_a|, a Lipschitz estimate, so a family of holomorphic functions whose derivatives have uniformly bounded modulus would also be (uniformly) equicontinuous. And since uniformly convergent sequences of holomorphic functions have holomorphic limits, we see that the limiting continuous function guaranteed by A-A is here also holomorphic.
Notice, please, we are using a sort of Mean Value inequality, which is certainly valid under these circumstances. We aren't using an equality form of MVT since that's not true.

It is amazing and wonderful and powerful that the requirements of Arzela-Ascoli can be combined in a simpler hypothesis for holomorphic functions.

Theorem (called Montel's Theorem) Suppose U is open, F is a collection of functions which are holomorphic on U, and for any K compact in U, sup_f in F||f||_K is finite (so the modulus of functions in F is bounded on any compact set). Then every sequence in F has a subsequence which converges uniformly on compact subsets.

The classical name for such F's is a normal family. These F's are subsets of the holomorphic functions whose closure is compact ("precompact" sets of holomorphic functions).

Proof U is the increasing union of compact sets K_n. We proved that ||f´||_K<=(1/s)||f||_{K_s} for sufficiently small positive s. So combine this with the previous observation, and we see that an F which is bounded on compact sets is also uniformly equicontinuous on compact sets. So A-A applies. And sigma-compactness also guarantees the existence of a (locally) uniformly convergent sequence of holomorphic functions in F. The limit function will be holomorphic.

(Not) finishing the proof of RMT: start with a well-chosen f on U. The hypothesis of simply connected insures that holomorphic square roots of non-zero holomorphic functions exist. This is used to create the initial f and then to continue with a sequence of f_n's. Montel's Theorem is used to guarantee a limit, and Hurwitz's Theorem (limits of 1-1 holomorphic functions are either constant or 1-1) to show that the limit is not constant. And the Schwarz Lemma essentially guarantees that the whole disc D₁(0) is filled up by the image of U. Please read one of the classical proofs. The text's version is in section 6.7.

For what follows we need one more classical result called the Schwarz Reflection Principle.

Schwarz Reflection (baby version) Suppose that U is open and connected in C, and f(z)=f(x+iy)=u(x,y)+iv(x,y) is holomorphic in U. If U^R is the collection of "reflected" z's in U (that is, all the z^bar's for z in U), then the function g(z)=[f(z^bar)^bar is holomorphic in U^R.

Why is this true? One way to think about this is to realize that f(z) near z₀ in U is the sum of a convergent power series. The mapping z-->z^bar is certainly continuous (although it equally certainly is not holomorphic). But then the series SUM_n=0^INFa_n(z-z₀)ⁿ becomes the series SUM_n=0^INF{a_n}^bar(z-{z₀}^bar)ⁿ upon the stated reflection, and this is a convergent power series in z, indeed.

Another way to verify holomorphicity is just to realize that g(z)=u(x,-y)-iv(x,-y). The several variable Chain Rule from calculus will show that g satisfies the Cauchy-Riemann equations.

Schwarz Reflection (adolescent version) Suppose that U is open and connected in H, the open upper halfplane (z's so that Im z>0). Also suppose that the boundary of U contains an interval I of the real line, R. Suppose that f is holomorphic in U and has real continuous boundary values on I: that is, lim_z-->xf(z)=Q(x) exists (x is in I and z is in U) and Q is continuous in I. Then define V to be the union of U and U^R (as defined before) and I also. Let g(z) be f(z) for z in U, [f(z^bar)]^bar for z in U^R, and Q(x) for x in I. Then g is holomorphic in V.

Proof Certainly the Baby Version of Schwarz Reflection shows that g is holomorphic in U and in U^R and we need to check what happens near I. But g is continuous in all of V. If we can show that the integral of g around a rectangle with sides parallel to the coordinate axes is 0, we are done (Morera!). But it's enough to consider a rectangle with one side on the x-axis. We can move it up slightly, and the integral is 0 (Cauchy!). This integral is continuous in the slight movement, so the desired rectangular contour integral must be 0.

This sort of argument was part of an earlier homework problem and a number of students recognized it. The hypotheses imply that the original mapping of f sort of sits on one side of the image R inside the image C. The reflection principle can be used in many contexts because both the domain and the image R's can be changed into many different curves by conformal mappings. This is used, for example, to analyze the automorphism groups of complicated domains such as an annulus, and can also be used to construct the elliptic modular function, a very important classical holomorphic function with uses in number theorem and hyperbolic geometry.

The adult version of the Schwarz Lemma weakens the hypothesis in the adolescent version in one wonderful way. We ask that the lim_z-->xf(z)=Q(x) still exist and be real, but no additional conditions on Q are needed. Notice that one consequence of the existence of the limit, using the adult version, is that the resulting Q turns out to be real analytic: certainly this is not obvious. It is not even obvious in the adolescent version, where there is an almost magical transition from "Q is continuous" to "Q is real analytic". This result is what's important in the very last discussion of the course. The proof of the adult verions uses the solution of the Dirichlet Problem and the consequence that continuous functions which have the mean value property must be harmonic. This is discussed in the textbook. See Theorem 7.5.2 in section 7.5.

The very last, very final, wonderful conclusion of the course is the Lewy example of a simple PDE with no solutions.

Also available is a direct discussion of the Mittag-Leffler and Weierstrauss Theorems, results guaranteeing the existence of holomorphic and meromorphic functions with strictly prescribed behavior. This was proofread by Mr. Castro and Ms. Naqvi, whose efforts I appreciated.

And, indeed, I appreciated the work of all the students in the course: thank you. Here is information about the final exam.