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Let $f$ be a continuous function on $[0,1]$. Suppose $\{f_n\}$ is a sequence of continuous functions on $[0,1]$ such that for any sequence $\{x_n\}$ in $[0,1]$, if $x_n \rightarrow x$ then $f_n(x_n) \rightarrow f(x)$.Show that $f_n$ converges uniformly to $f$ on $[0,1]$.

My attempt $:$

It is clear that for each $x \in [0,1]$ there exists a sequence $\{x_n\}$ in $[0,1]$ such that $x_n \rightarrow x$.Because take any $x \in (0,1)$. Let $1-x = \delta>0$.then by Archimedian property there exists a natural number $m$ such that $\frac {1} {m} < \delta$.Now take a sequence $\{x_n\}$ where $x_n = x + \frac {1} {n+m-1}$ , $n \in \mathbb N$.Then clearly $\{x_n\}$ is a sequence in $[0,1]$ converges to $x$.Again if $x=0$ (resp. $x=1$) then we take $x_n = \frac {1} {n}$ , $n \in \mathbb N$ (resp. $x_n = 1-\frac {1} {n}$ , $n \in \mathbb N$) for having a sequence $\{x_n\}$ in $[0,1]$ converges to $0$ (resp. $1$).

Now let us take $x \in [0,1]$ arbitrarily.So according to the above argument $\exists$ a sequence $\{x_n\}$ in $[0,1]$ which converges to $x$.Since $f_n$ is continuous for all $n \in \mathbb N$ so by sequential criterion for continuity the sequence $\{f_n(x_n)\}$ should converge to $f_n(x)$.Since a convergent sequence can have at most one limit we have $f_n(x) = f(x)$ for all $n \in \mathbb N$ (since it is given that if $x_n \rightarrow x$ then $f_n(x_n) \rightarrow f(x)$ ).Since $x \in [0,1]$ is arbitrary we have $f_n(x) = f(x)$ for all $x \in [0,1]$ and for all $n \in \mathbb N$.Hence for all $x \in [0,1]$ the term $|f_n(x)-f(x)|$ can be made less than any positive $\epsilon$ for all $n \ge 1$.This proves that $f_n(x) \rightarrow f(x)$ as $n \rightarrow \infty$ uniformly on $[0,1]$.

Is the above attempt correct at all?Please verify it.

Thank you in advance.

Arnab Chattopadhyay.
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1 Answers1

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Your proof is not correct. When you say that there is a sequence $(x_n)_n$ in $[0,1]$ which converges to $x$, this could be the constant sequence with $x_n=x$. Then the second part of your proof would imply that pointwise convergence is equivalent to uniform convergence which is not true.

Alternative approach. If $(f_n)_n$ does not converge uniformly to $f$ on $[0,1]$, then, by definition, there is $\epsilon>0$ and a subsequence $(n_k)_k$ such that $$\sup_{x\in [0,1]}|f_{n_k}(x)-f(x)|>\epsilon.$$ By $\sup$ definition, for any $k$ there is $x_{n_k}\in [0,1]$ such that $$|f_{n_k}(x_{n_k})-f(x_{n_k})|>\frac{\epsilon}{2}\tag{1}.$$ Note that $[0,1]$ is compact, and by Bolzano–Weierstrass theorem, there is a subsequence $(x_{n_{k_j}})_j$ (sorry for the notation) which converges to $x\in [0,1]$. Hence, by the hypothesis, $$f_{n_{k_j}}(x_{n_{k_j}})\to f(x)\quad\mbox{and}\quad f(x_{n_{k_j}})\to f(x)\implies |f_{n_{k_j}}(x_{n_{k_j}})-f(x_{n_{k_j}})|\to 0$$ which contradict $(1)$.

Robert Z
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