# another proof of Dini’s theorem

Besides, $f_{n}(x)\geq f_{n+1}(x)\quad\forall x\in K,\forall n$.

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

Proof

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

 $\exists\varepsilon>0\;\text{such that }\forall m\in N\;\exists\;n_{m}>m,\;x_{m% }\in K\;\text{such that }\quad\left|{f_{n_{m}}(x_{m})-f(x_{m})}\right|\geq\varepsilon.$

So,

 $\begin{array}[]{l}\text{For }\;m=1\;\exists\;n_{1}>1\;,\;x_{1}\in K\;\text{% such that }\;\left|{f_{n_{1}}(x_{1})-f(x_{1})}\right|\geq\varepsilon\\ \exists\;n_{2}>n_{1}\;,\;x_{2}\in K\;\text{such that }\;\left|{f_{n_{2}}(x_{2}% )-f(x_{2})}\right|\geq\varepsilon\\ \vdots\\ \exists\;n_{m}>n_{m-1}\;,\;x_{m}\in K\;\text{such that }\;\left|{f_{n_{m}}(x_{% m})-f(x_{m})}\right|\geq\varepsilon\\ \end{array}$

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

 $x_{m_{j}}\longrightarrow x\in K$

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

Let $j_{0}\;\text{such that }\;j\geq j_{0}\Rightarrow\left|{f_{n_{m_{j}}}(x)-f(x)}% \right|<\raise 3.01pt\hbox{\varepsilon}\!\mathord{\left/{\vphantom{% \varepsilon 4}}\right.\kern-1.2pt}\!\lower 3.01pt\hbox{4}$, which exists due to the punctual convergence of the sequence. Then, particularly, $\left|{f_{n_{m_{j_{o}}}}(x)-f(x)}\right|<\raise 3.01pt\hbox{\varepsilon}\!% \mathord{\left/{\vphantom{\varepsilon 4}}\right.\kern-1.2pt}\!\lower 3.01pt% \hbox{4}$.

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 $f_{n}(y)\geq f_{n+1}(y)\quad\forall y\in K,\forall n$) it’s easy to see that

 $f_{n}(y)\geq f(y)\quad\forall y\in K,\forall n$

Then, $f_{n_{m_{j}}}(x_{m_{j}})-f(x_{m_{j}})\geq\varepsilon\;\forall j$. And also the hypothesis implies

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

So, $j\geq j_{0}\Rightarrow f_{n_{m_{j_{0}}}}(x_{m_{j}})\geq f_{n_{m_{j}}}(x_{m_{j}% }),$ which implies

 $\left|{f_{n_{m_{j_{0}}}}(x_{m_{j}})-f(x_{m_{j}})}\right|\geq\varepsilon$

Now,

 $\left|{f_{n_{m_{j_{0}}}}(x_{m_{j}})-f(x)}\right|+\left|{f(x_{m_{j}})-f(x)}% \right|\geq\left|{f_{n_{m_{j_{0}}}}(x_{m_{j}})-f(x_{m_{j}})}\right|\geq% \varepsilon\;\;\forall j\geq j_{0}$

and so

 $\left|{f(x_{m_{j}})-f(x)}\right|\geq\varepsilon-\left|{f_{n_{m_{j_{0}}}}(x_{m_% {j}})-f(x)}\right|\;\;\forall j\geq j_{0}.$

On the other hand,

 $\left|{f_{n_{m_{j_{0}}}}(x_{m_{j}})-f(x)}\right|\leq\left|{f_{n_{m_{j_{0}}}}(x% _{m_{j}})-f_{n_{m_{j_{0}}}}(x)}\right|+\left|{f_{n_{m_{j_{0}}}}(x)-f(x)}\right|$

And as $f_{n_{m_{j_{0}}}}$ is continuous, there is a $j_{1}$ such that

 $j\geq j_{1}\Rightarrow\left|{f_{n_{m_{j_{0}}}}(x_{m_{j}})-f_{n_{m_{j_{0}}}}(x)% }\right|<\raise 3.01pt\hbox{\varepsilon}\!\mathord{\left/{\vphantom{% \varepsilon 4}}\right.\kern-1.2pt}\!\lower 3.01pt\hbox{4}$

Then,

 $j\geq j_{1}\Rightarrow\left|{f_{n_{m_{j_{0}}}}(x_{m_{j}})-f(x)}\right|\leq% \left|{f_{n_{m_{j_{0}}}}(x_{m_{j}})-f_{n_{m_{j_{0}}}}(x)}\right|+\left|{f_{n_{% m_{j_{0}}}}(x)-f(x)}\right|<\raise 3.01pt\hbox{\varepsilon}\!\mathord{\left/% {\vphantom{\varepsilon 2}}\right.\kern-1.2pt}\!\lower 3.01pt\hbox{2},$

which implies

 $\left|{f(x_{m_{j}})-f(x)}\right|\geq\varepsilon-\left|{f_{n_{m_{j_{0}}}}(x_{m_% {j}})-f(x)}\right|\geq\raise 3.01pt\hbox{\varepsilon}\!\mathord{\left/{% \vphantom{\varepsilon 2}}\right.\kern-1.2pt}\!\lower 3.01pt\hbox{2}\;\;% \forall j\geq\max\left({j_{0},j_{1}}\right).$

Then, particularly, $f(x_{m_{j}})_{j}$ does not converge to $f(x)$. QED.

Title another proof of Dini’s theorem AnotherProofOfDinisTheorem 2013-03-22 14:04:37 2013-03-22 14:04:37 gumau (3545) gumau (3545) 11 gumau (3545) Proof msc 54A20