Converges Uniformly

Let , and . We say converges uniformly to if:

We write uniformly.

Note that if pointwise means:

So we just swapped two quantifiers. It is now easy to see that uniform implies pointwise.

Visually, all functions need to be in this -neighbourhood of :

\begin{document}
\begin{tikzpicture}[domain=0:4]
  \draw[very thin,color=gray] (-0.1,-1.1) grid (3.9,1.7);
 
  \draw[->] (-0.2,0) -- (4.2,0) node[right] {$x$};
  \draw[->] (0,-1.2) -- (0,1.9) node[above] {$f$};
 
  %\draw[color=red]    plot (\x,\x)             node[right] {$f(x) =x$};
  % \x r means to convert '\x' from degrees to _r_adians:
  \draw   plot (\x,{sin(\x r)});
  \draw[dashed] plot(\x, {sin(\x r) + 0.1});
  \draw[dashed] plot(\x, {sin(\x r) - 0.1});
  % \draw[color=orange] plot (\x,{0.05*exp(\x)}) node[right] {$f(x) = \frac{1}{20} \mathrm e^x$};
\end{tikzpicture}
\end{document}

Theorem CTS

Suppose each is Continuous and uniformly. Then is CTS.

Proof

Let and . By uniform convergence, find s.t.

In particular,

By continuity of at , we can find s.t.

Now let s.t. . Have

Hence is CTS.

This is called a -proof.

Theorem INT

Let the domain . Suppose uniformly and each is Riemann Integrable. Then is also Riemann Integrable and .

Proof

Take large enough so that . Note that is bounded (by ). Hence is bounded by . Now for any dissection , and are defined. Pick . Hence find big enough s.t. and find a dissection s.t. . Furthermore, can find that and similar for

Now use triangle inequality on to find a proof.

Finally, have:

Hence .

Corollary

Suppose is a sequence of functions whose partial sums converge uniformly to . Then:

Theorem Diff

Let be . Assume that the sequence of partial sums of converges uniformly and that there is some such that:

converges. Then the sequence of partial sums converges uniformly. Furthermore, the limit is and:

Proof sketch

Define where is the uniform limit of partial sums of . Note that by FTA, is differentiable and it’s derivative is . So we just need to prove that partial sums of converge uniformly to . Let . Find s.t. for any we have AND . By FTA we have . So: