another proof of Dini’s theorem

This is the version of the Dini’s theorem I will prove:Let $K$ be a compact metric space and ${\left(f_n\right)}_{n\in N}\subset C(K)$ which converges pointwise to $f\in C(K)$.

Besides, $f_n(x)\ge f_{n+1}(x)\mathit{\hspace{1em}}\forall x\in K,\forall n$.

Then ${\left(f_n\right)}_{n\in N}$ converges uniformly in $K$.

Proof

Suppose that the sequence does not converge uniformly. Then, by definition,

$$\exists \epsilon >0\text{such that }\forall m\in N\exists n_m>m,x_m\in K\text{such that }\mathit{\hspace{1em}}\left|f_{n_m}(x_m)-f(x_m)\right|\ge \epsilon .$$

So,

$$\begin{array}{c}\text{For }m=1\exists n_1>1,x_1\in K\text{such that }\left|f_{n_1}(x_1)-f(x_1)\right|\ge \epsilon \hfill \\ \exists n_2>n_1,x_2\in K\text{such that }\left|f_{n_2}(x_2)-f(x_2)\right|\ge \epsilon \hfill \\ \mathrm{⋮}\hfill \\ \exists n_m>n_{m-1},x_m\in K\text{such that }\left|f_{n_m}(x_m)-f(x_m)\right|\ge \epsilon \hfill \end{array}$$

Then we have a sequence ${\left(x_m\right)}_m\subset K$ and ${\left(f_{n_m}\right)}_m\subset {\left(f_n\right)}_n$is a subsequence of the original sequence of functions. $K$ is compact, so there is a subsequence of ${\left(x_m\right)}_m$which converges in $K$, that is, ${\left(x_{m_j}\right)}_j$ such that

$$x_{m_j}⟶x\in K$$

I will prove that $f$ is not continuous in $x$ (A contradiction with one of the hypothesis).

To do this, I will show that $f{\left(x_{m_j}\right)}_j$ does not converge to $f(x)$, using above’s $\epsilon$.

Let ,which exists due to the punctual convergence of the sequence. Then,particularly, .

Note that

$$\left|f_{n_{m_j}}(x_{m_j})-f(x_{m_j})\right|=f_{n_{m_j}}(x_{m_j})-f(x_{m_j})$$

because (using the hypothesis $f_n(y)\ge f_{n+1}(y)\mathit{\hspace{1em}}\forall y\in K,\forall n$) it’s easy to see that

$$f_n(y)\ge f(y)\mathit{\hspace{1em}}\forall y\in K,\forall n$$

Then, $f_{n_{m_j}}(x_{m_j})-f(x_{m_j})\ge \epsilon \forall j$. And also the hypothesis implies

$$f_{n_{m_j}}(y)\ge f_{n_{m_{j+1}}}(y)\forall y\in K,\forall j$$

So, $j\ge j_0\Rightarrow f_{n_{m_{j_0}}}(x_{m_j})\ge f_{n_{m_j}}(x_{m_j}),$ which implies

$$\left|f_{n_{m_{j_0}}}(x_{m_j})-f(x_{m_j})\right|\ge \epsilon$$

Now,

$$\left|f_{n_{m_{j_0}}}(x_{m_j})-f(x)\right|+\left|f(x_{m_j})-f(x)\right|\ge \left|f_{n_{m_{j_0}}}(x_{m_j})-f(x_{m_j})\right|\ge \epsilon \forall j\ge j_0$$

and so

$$\left|f(x_{m_j})-f(x)\right|\ge \epsilon -\left|f_{n_{m_{j_0}}}(x_{m_j})-f(x)\right|\forall j\ge j_0.$$

On the other hand,

$$\left|f_{n_{m_{j_0}}}(x_{m_j})-f(x)\right|\le \left|f_{n_{m_{j_0}}}(x_{m_j})-f_{n_{m_{j_0}}}(x)\right|+\left|f_{n_{m_{j_0}}}(x)-f(x)\right|$$

And as $f_{n_{m_{j_0}}}$ is continuous, there is a $j_1$such that

Then,

which implies

$$\left|f(x_{m_j})-f(x)\right|\ge \epsilon -\left|f_{n_{m_{j_0}}}(x_{m_j})-f(x)\right|\ge \epsilon \mathrm{/}2\forall j\ge \max (j_0,j_1).$$

Then, particularly, $f{(x_{m_j})}_j$ does not converge to $f(x)$. QED.