Uniform convergence on subintervals

Question

I have just started learning pointwise convergence and uniform convergence, and I am facing some confusion between two statements.

1. (Proven to be true) Let $f_n : \Bbb{N} \to \Bbb{R}$ be a sequence of functions which converges pointwise to a limit $f:\Bbb{N} \to \Bbb{R}$. This implies that $f_n \to f$ uniformly on {$\text{$1,...,N$}$} for any N $\epsilon$ $\Bbb{N}$

Now I have another statement:

2. Let M be a fixed positive integer M and let ${f_n:[0,M] \to \Bbb{R}}$ be a sequence of functions. Suppose that $f_n \to f$ pointwise on $[0,M]$ and that $f_n \to f$ uniformly on each subinterval $[k-1,k]$ for each $k=1,...,M$. Prove that $f_n \to f$ uniformly on $[0,M]$.

I cannot figure out how the condition of uniform convergence on subintervals changes the question. Doesn't the first statement proves the second statement? Hope someone can enlighten me on this.

Thanks in advance!

Edit: I now have an idea the difference between these two statements. Can anybody guide me on proving the second statement then?

Edit: I constructed a proof for the second statement:

By assumption, for any $\varepsilon > 0$, there exists $N_k(\varepsilon)$ such that $$|f_n(x)-f(x)|\le \varepsilon$$ for all $n \ge N_k(\varepsilon)$

We choose $N^*(\varepsilon)=\max\{N_1,...,N_{M-1}\}$.

Hence, for any $\varepsilon > 0$, there exists $N^*(\varepsilon)$ such that $$|f_n(x)-f(x)|\le \sup_{x\in[0,M]}|f_n(x)-f(x)|$$ $$=\max\{\sup_{x\in[0,1]}|f_n(x)-f(x)|,...,\sup_{x\in[M-1,M]}|f_n(x)-f(x)|\}$$ $=\varepsilon$ for all $n \ge N^*(\varepsilon)$

We conclude that $f_n \to f$ uniformly on $[0,M]$

Is this proof logical?

Answer 1

The first statement does not prove the second. Take $f_n(x)=x^n$ then it converges uniformly in $\{0,1\}$ but it does not converge uniformly in $[0,1]$.

As regards a proof of the first and the second statement, basically we use the same fact: if we have uniform convergence in $A$ and in $B$ then we have uniform convergence in $A\cup B$ (which implies that we can take the union of any finite number of sets). This follows from the property: $$\sup_{x\in A\cup B}|f_n(x)-f(x)|=\max\left(\sup_{x\in A}|f_n(x)-f(x)|, \sup_{x\in B}|f_n(x)-f(x)|\right).$$