Let $f_n$ be a sequence of continuous, differentiable functions on $[0,1]$ converging pointwise to $f$ such that:
- $f_n\to f$ uniformly on $[0,\lambda]$ for all $0<\lambda<1$
- $f$ continuous on $[0,1]$, differentiable on $[0,1)$
Is it true that $f_n\to f$ uniformly on $[0,1]?$ What if $f_n(x)=\sum_{k=0}^n a_k x^k$ for some convergent $\sum a_k?$
Edit: first part solved, thanks guys. Any help on the second part would be greatly appreciated.
Initial try:
I think the answer is yes, but I am having trouble proving it. I tried this:
Taking $\epsilon>0$ we can find $\delta>0$ such that $|1-x|<\delta\Rightarrow |f(1)-f(x)|<\epsilon/3$. Take $\xi\in (1-\delta, 1]$. It would be enough to show that, for some $N$, $|f_n(x)-f(x)|<\epsilon$ for all $x\in [\xi,1]$ and $n\geq N$. We could choose $N$ so that $\forall n>N:$
$$|f_n(x)-f(x)|\leq \underbrace{|f(x)-f(1)|}_{<\epsilon/3}+\underbrace{|f(1)-f_n(1)|}_{<\epsilon/3}+\underbrace{|f_n(1)-f_n(x)|}_{<\epsilon/3 ?}<\epsilon ?$$
I can't see how to use the rest of the information to restrict the last modulus, or whether this is even a good route to take.
Counterexample: Let $\eta$ be a nontrivial smooth function supported in $[0,1]$. Then
$$f_n(x) = \eta( 1- nx)$$
has support in $[1-\frac 1n, 1]$. $f_n$ converges uniformly to the zero function on $[0,\lambda]$ for all $\lambda <1$, and converge pointwisely to the zero function on $[0,1]$. But the convergence is not uniform.
However, if
$$f_m(x) = \sum_{n=0}^m a_n x^n,\ \ \ f(x) = \sum_{n=0}^\infty a_n x^n$$
and $\sum a_n$ exists, then the assertion is true.
Proof: Let $s_n = a_0+ a_1 +\cdots a_n, s_{-1} = 0$. Note that $s_n \to s = \sum a_n$. Now we have
$$f_m(x) = \sum_{n=0}^m a_nx^n = \sum_{n=1}^m (s_n - s_{n-1})x^n = (1-x)\sum_{n=0}^{m-1} s_n x^n + s_m x^m$$
This means
$$\begin{split}|f_m(x) - f_m(1)| &= \left| (1-x)\sum_{n=0}^{m-1} s_n x^n + s_m x^m - s_m \right|\\ &=\left| (1-x)\sum_{n=0}^{m-1} s_n x^n - s_m(1-x) \sum_{n=0}^{m-1} x^n\right| \\ &= (1-x) \left| \sum_{n=0}^{m-1} (s_n -s_m) x^n \right|. \end{split}$$
Now as $s_n \to s$, then $\{s_n\}$ is Cauchy and so there is $K$ so that $|s_n - s_m| < \epsilon/6$ for all $m, n \ge K$.
Now for all $m \ge K$,
$$\begin{split} (1-x) \left| \sum_{n=0}^{m-1} (s_n -s_m) x^n \right| &=(1-x) \left| \sum_{n=0}^{K} (s_n -s_m) x^n + \sum_{n=K}^{m-1} (s_n -s_m) x^n\right| \\ &\le (1-x) M_K + \frac{\epsilon }{6}(1-x) \sum_{n=K}^{m-1} x^n \\ &\le (1-x) M_K + \frac\epsilon 6(1-x) \sum_{n=0}^{\infty} x^n \\ &= (1-x) M_K + \frac\epsilon 6 \end{split}$$
for all $x\in [0,1]$. (Note that the argument does not work for $x=1$, but the inequality is satisfied trivially in that case). In particular, if $x$ is close to one so that $|1-x |< \frac{\epsilon}{6M_K}$, then
$$|f_m(x) - f_m(1)| <\frac{\epsilon}{3}$$
for all $m\ge K$ and $x$ close to $1$. This is precisely what you need to show the uniform convergence.
Remark: The above proof is a small modification of the proof of Abel's theorem that I found in Rudin's PMA.