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Let $f_n$ be a sequence of function from [0,1] to $\mathbb{R}$ pointwise converge to $f$. Supposed for all sequence $(x_n)$ in [0,1] converge to $x$ in [0,1], then $f_n(x_n)$ converge to $f(x)$. Show that $f_n$ is uniform convergent on [0,1]

I have seen a familiar exercise to this Show that the sequence of functions {fn} convereges uniformly to f on [0,1] by the given condition.

But there is a different here which is in the exercise I am working on, we are given that $f_n$ is pointwise convergence instead of $f$ being continue. And thus I can not prove this one.

Thanks for all the helping !

user635988
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1 Answers1

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Assume wlog that $f=0.$

By contradiction, assume that the convergence is not uniform, i.e. there exists an $\varepsilon>0$ such that $$\forall N\in\mathbb N\quad\exists n_N\ge N\quad\exists y_N\in[0,1]\quad|f_{n_N}(y_N)|\ge\varepsilon$$ and wlog, $(n_N)_N$ is strictly increasing.

By compacity of $[0,1]$, $(y_N)_N$ admits a convergent subsequence, $$y_{N_k}\to x\quad(k\to+\infty).$$

Let $(x_n)_n$ be any sequence in $[0,1]$ such that $$\left(\forall k\in\mathbb N\quad x_{n_{N_k}}=y_{N_k}\right)\quad \text{and}\quad x_n\to x.$$ (Such a sequence is easy to construct, e.g. letting the sequence be constant between two consecutive prescribed values.)

Then, $(f_n(x_n))_n$ does not converge to $f(x)=0$, since its subsequence $(f_{n_{N_k}}(x_{n_{N_k}}))_k$ doesn't.

Anne Bauval
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