Uniform convergence, functions,sequences,subsequences

Question

Let $f_{n}$ be a sequence of continuous functions with $f$ continuous on $\left[0,1\right]$. Prove that if $x_{n}$ is a sequence in $\left[0,1\right]$ such that $\lim_{n\to\infty}x_{n}=x$ for some $x\in \left[0,1\right]$ implies $\lim_{n\to\infty}f_{n}(x_{n})=f(x)$.Then $f_{n}$ converges uniformly to $f$ on $\left[0,1\right]$.

I tried to prove by contrapositive. If $f_{n}$ does not converge to $f$ uniformly on $\left[0,1\right]$, then there exists an $\epsilon_{0}>0$ such that, for all $N>0$, there exists an $n>N$ and $x\in\left[0,1\right]$ such that $|f_{n}(x)-f(x)|\ge\epsilon_{0}$.

For example for $N=1$, there exists an $x_{1}\in \left[0,1\right]$ and $n_{1}>1$ such that $|f_{n_1}(x_{1})-f(x_1)|\ge \epsilon_{0}$.We can do this for any $N=k$ and choose $n_{k}>k$, construct a sequence $x_{k} \in \left[0,1\right]$ such that $|f_{n_k}(x_{k})-f(x_{k})|\ge \epsilon_{0}$. Since every bounded sequence has a convergent subsequence by Bolzano Weierstrass theorem there exists a subsequence $x_{k_{l}}$ of $x_{k}$ such that $\lim_{k_{l}\to\infty}x_{k_{l}}=x$ for some $x \in \left[0,1\right]$. Then $1)$ $\lim_{k_{l}\to\infty} f_{k_{l}}(x_{k_{l}})=f(x)$. Then there exists a contradiction,

$$|f(x_{k_{l}})-f_{n_{k_{l}}}(x_{k_{l}})|<|f(x_{k_{l}})-f(x)|+|f(x)-f_{k_{l}}(x_{k_{l}})|+|f_{k_{l}}(x_{k_{l}})-f_{n_{k_{l}}}(x_{k_{l}})|$$

Now it can be clearly seen that $f_{n}$ converges to $f$ pointwise and although left side is greater than $\epsilon_{0}$ right side can be made less than $\epsilon_{0}$ by using pointwise convergence Cauchyness the result $1)$ and the continuity of the function $f$.Is everything fine?Is my proof well?

Answer 1

Two issues in your work.

(1) Small issue:

For example for $N=1$, there exists an $x_{1}\in \left[0,1\right]$ and $n_{1}>1$ such that $|f_{n_1}(x_{1})-f(x_1)|\ge \epsilon_{0}$.We can do this for any $N=k$ and choose $n_{k}>k$, construct a sequence $x_{k} \in \left[0,1\right]$ such that $|f_{n_k}(x_{k})-f(x_{k})|\ge \epsilon_{0}$.

Subsequence needs to be ascending on its index, hence, for $N=2$, you need to pick $n_2$, such that $n_2>\max(2, n_1)$. Keep going, for $N=k$, pick $n_k$, such that $n_k>\max(k, n_{k-1})$.

(2) Big issue:

$$|f(x_{k_{l}})-f_{n_{k_{l}}}(x_{k_{l}})|<|f(x_{k_{l}})-f(x)|+|f(x)-f_{k_{l}}(x_{k_{l}})|+|f_{k_{l}}(x_{k_{l}})-f_{n_{k_{l}}}(x_{k_{l}})|$$

How do you justify the last term $|f_{k_{l}}(x_{k_{l}})-f_{n_{k_{l}}}(x_{k_{l}})|<\epsilon$ ? For simple, we define $p=k_l$, then the last term is converted to

$$|f_{p}(x_{p})-f_{n_{p}}(x_{p})|\underset{?}<\epsilon$$

For example, let $f_p(x)=x^p, x_p=1-\frac1p$, then $$\lim_{p\to\infty}f_p(x_p)=\lim_{p\to\infty}\left(1-\frac1p \right)^p= e^{-1}$$

In your construction, there is no "complete" control for the relation between $n_p$ and $p$. So if we set $n_p=p^2$

$$\lim_{p\to\infty}f_{n_{p}}(x_{p})=\lim_{p\to\infty}\left(1-\frac1p \right)^{p^2}=0$$

So the last term $|f_{k_{l}}(x_{k_{l}})-f_{n_{k_{l}}}(x_{k_{l}})|>\frac12e^{-1}$, which is not approaching to zero and you can't say it is a contradiction.