\section{Derivatives of Limits}

\subsection{Intro}
The following is \emph{careful} analysis.
That is, if you state the theorems in the \emph{right} generality, then the
proofs are \emph{immediate}.

Much of the confusion comes from over-using uniform convergence:
uniform convergence implies $L^1$ convergence (on compact interval)
and implies that limit of continuous is continuous.

As this is an immediate application of the FTC, which has stronger and
weaker forms, it admits a strong form, which is how we state it, and a
weak form, which is how it's usually stated.

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\subsection{Main theorem}
Let ID be (everywhere differentiable functions with integrable derivative).
\begin{thm}[Strong, analytic version]
If:
\begin{itemize}
 \item $f_n$ a sequence of ID functions
 \item $f'_n \to g$ in $L^1$
 \item $f_n$ converges pointwise at a \emph{single} point $a$;
\end{itemize}
then: $f_n$ converges uniformly, to
$$f(x) = \lim f_n(a) + \int_a^x g.$$
\end{thm}
\begin{proof}
Briefly: apply FTC (second form) and take limits.

Formally:
as $f'_n$ is integrable, $f_n(x) = f_n(a) + \int_a^x f'_n$.

Then
$$\abs{f(x)-f_n(x)} \leq \abs{f(a)-f_n(a)} + \int_a^x (g-f'_n)$$
so
$$\norm{f(x)-f_n(x)}_\infty \leq \abs{f(a)-f_n(a)} + \norm{g-f'_n}_1$$
Since the RHS goes to $0$, so does the LHS.
\end{proof}
Note that $f$ is continuous, but not necessarily differentiable.

\begin{cor}[Corollary of proof]
If (in addition) $g$ is continuous, then $f'=g$ (which $=\lim f_n$, in $L^1$).

If (even more) $f'_n \to g$ pointwise, then $f'(x)=\lim f'_n(x)$.
\end{cor}
\begin{proof}
For the first statement, apply the first form of the FTC.

The second statement is immediate.
\end{proof}

\begin{cor}[Weak, analytic version]
If:
\begin{itemize}
 \item $f_n$ a sequence of $C^1$ functions;
 \item $f'_n \to g$ uniformly on a compact set;
 \item $f_n$ converges pointwise \emph{at a single point} $a$;
\end{itemize}
then: $f_n$ converges uniformly, say to $f$, and then $f'=g$
and $f'(x) = \lim f'_n(x)$.
\end{cor}
\begin{proof}
Just check the conditions:\\
Continuous functions are integrable, and uniform convergence on a
compact set implies $L^1$ convergence.
Uniform convergence implies pointwise, so $f_n \to g$ pointwise.
Further, $g$ is automatically continuous, as it's the uniform limit of
continuous.
\end{proof}
Note that this is how it's often stated and proved; you only need a
weak form of the FTC to prove this (assuming your derivatives are
continuous), and the proof is essentially as we gave it, but all the
implications of uniformity are mixed in with the actual analysis.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Interpretation}
What's the \emph{meaning} of this?
Briefly, $L^1$ convergence of derivatives, plus pointwise convergence
at a single point, both very weak conditions, imply \emph{uniform}
(IE, $L^\infty$) convergence of the original sequence of functions, which
is very strong.

Here's some examples to show why all the hypotheses are needed:
\begin{itemize}
 \item if $f'$ are not integrable, we can't even talk about $L^1$
    convergence: that's a requirement to state the theorem
 \item we need convergence of $f_n$ at -some- point, otherwise we can get
   $f_n(x) \equiv n$. Then $f'_n \equiv 0$, which certainly converges,
   but $f_n$ doesn't converge to anything.
 \item if we don't have pointwise convergence of $f'_n$,
   then we needn't have $f'(x)=\lim f'_n(x)$, for instance
   $\cos(n^2x)/n$ converges uniformly to $0$, but the derivative
   doesn't converge is $L^1$ or pointwise at any points (except
   $k\pi$).
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Rudin note}
If we don't assume that the $f'_n$ are integrable,
but we do assume uniform convergence of $f'_n$,
then the strong conclusion
  ($f_n \to f$ uniformly, $f'(x) = \lim f'_n(x)$)
holds.

This is much trickier analysis, involving the MVT.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Formal/Geometric}
Above, we did actual analysis.
However, the form in which geometers use the theorem is:
\begin{thm}[Geometric version]
$D\from C^1 \to C^0$ is continuous.
IE, if $f_n \to f$ in $C^1$, then $f'_n \to f'$ in $C^0$:
derivatives commute with limits, if the topologies are correct.
\end{thm}

In this form it's tautological:
the $C^1$ topology is defined as the subspace topology of
$$C^1 \to C^0 \oplus C^0$$
by $f \mapsto (f,f')$, where $C^0$ has the uniform topology.
Then $D\from C^1 \to C^0$ is exactly
$C^1 \to C^0 \oplus C^0 \to C^0$ by
$f \mapsto (f,f') \mapsto f'$, which is continuous by construction.

Similarly, $C^k$ has the weak topology of
$$C^k \to C^{k-1} \oplus C^{k-1}$$
by $f \mapsto (f,f')$ (I think...).

IE, the $C^k$ topology is the weak topology such that
$i,D\from C^k \to C^{k-1}$ (inclusion and differentiation)
are continuous (I think...).

Further, we actually get a norm on $C^k$ by taking the sum or max of
the norms of $f,\dots,f^{(k)}$ -- which is okay if we're compact.
If we're just locally compact, then we get a family of seminorms
(indexed by the compact subsets), etc.

Why do this? Why not just use the uniform topology on $C^k$?
Because $C^1$ is not complete in the $C^0$ topology
(equiv., $C^1 \subset C^0$ is not closed in the $C^0$ topology),
and more generally $C^k$ is not complete in the $C^{k-1}$ topology.

That is, consider a family of smooth function that have a uniform
limit that's not differentiable, like $f_a(x) = \sqrt{x^2 + a^2}$:
these are hyperbolae that converge uniformly to $f_0=\abs{x}$.
Note that the derivatives converge in $L^1$, but to the Heaviside step
function, which is not continuous at $0$, hence $f_0$ needn't be
differentiable at $0$, in agreement with reality.

Spivak has a nice illustration in Chapter 24.

What this is showing is that the corner is getting sharper and
sharper, even though the function isn't moving much:
just because your norm is low doesn't mean your \emph{derivative} is
low: hence $D\from C^1 \to C^0$ isn't bounded (hence not continuous)
in the $C^0$ topology.
(It's also cute to look at the derivatives, which give a sequence
converging $L^1$ and pointwise but not $L^\infty$ to the step function.)

OTOH if we require $f_n$ and $f'_n$ (etc., up to $f^{(k)}_n$) to converge,
then obviously $C^k$
is complete: that is, if $f_k$ is a Cauchy sequence in the $C^k$
topology, then take a subsequence such that all of the derivatives
converge (just need the $k$th derivative to converge, and the rest
to converge at some point), and then FTC/Taylor shows that you
converge to a $C^k$ function.

So the $C^k$ topology/norm is so that:
\begin{itemize}
 \item the space is complete/Banach
 \item differentiation is continuous
\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

FIXME: below here is a mess, and probably can just be cut.


\begin{lem}[formal]
If $f_n$ is a sequence of $C^1$ functions
such that $f_n \to f$ uniformly and $f'_n \to g$ uniformly,
then $g=f'$.

IE, when convergence of the functions and the derivatives is uniform,
then derivatives commute with limits.

Tersely: derivatives commute with $C^1$ limits of $C^1$ functions.

Most tersely: $D\from C^1 \to C^0$ is continuous.
\end{lem}
Note however that they do -not- commute with $C^0$ limits of $C^1$
functions.
\begin{proof}
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
URK: this isn't general/careful enough: see above.
The big/general statement is:
\begin{thm}
If $f_n$ is a sequence of $C^1$ functions on a compact set, and
 - $f_n$ converges -pointwise- to $f$
 - $f'_n$ converges UNIFORMLY to -something- $g$,
 then:
 a: $f_n$ converges UNIFORMLY to $f$
 b: $g=f'$, IE, derivatives commute with limits.

More generally, we can replace compact/uniform by ``uniformly on
compacts''.
\end{thm}

- the -uniform- limit of differentiable functions is differentiable
  (and the derivative is the limit of the derivatives:
   $f' = \lim f'_n$)
  -if- the derivatives converge to -some- function $g$
    ** this is a weird swerve, that's easy to miss **
    this is because $C^1$ is not complete in the $C^0$ topology
    ...and is trivially complete in the $C^1$ topology.
    (note counter example: $\cos(n^2x)/n$)
    (also example of smooth functions with limit not differentiable)
  proof:
    integrate the derivatives

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  NB: here's another proof: saying the derivatives converge to some
  function $g$ (and the functions converge to $f$) says that it's
  Cauchy in the $C^1$ topology, hence convergent
  ...and the $C^1$ topology is exactly the pullback of the $C^0 \oplus
  C^0$ topology under $f\mapsto (f,f')$, so by continuity (which is
  true by definition of pullback/subspace topology!), this theorem is
  true.
*** IE, with the right topology etc., this theorem is obvious ***
 
 I proved the -second- part of the theorem,
 namely: if $(f_n,f'_n)$ converges uniformly to $(f,g)$ (in $C^0$)
   then $f'=g$.
   ...and this is -obvious-: $\set{(f,f')}$ is a closed subset of
   $C^0 \oplus C^0$ b/c it's a kernel.
 
 the -first- part (a) seems -much- trickier and to require some
 topology.

  NB: this is -actually- saying something about different norms on
  function spaces (and you see that you only need $f'_n$ to be bounded norm):
  the map $C^0 \to C^1$ (i.e., change the topology on $C^1$) is not
   continuous/bounded
    (as the example $\cos nx$ shows; the above counterexample is just
    a massaging of this)
    (the other way map is obviously continuous)
hmm...here's a way of saying it:
it's just saying that ker $[D -I]$ (i.e., $(f,g) \mapsto f'-g$) is
closed in $C^1 \oplus C^0$ (with $C^0$ aka norm topology on both),
which should be clear: kernels of continuous maps are closed
...hey, but why is this map continuous?
  ...this is tricky...
...okay, it's not true for good, abstract reasons:
  you want to define this function $[D -I]$ on $C^1 \oplus C^0$
  ...and $C^1$ is -not- a Banach space in the $C^0$ topology!!!


