If $\lim_n f_n(x_n)=f(x)$ for every $x_n \to x$ then $f_n \to f$ uniformly on $[0,1]$?

Question

This is a self-posed question, so I do not know the answer and I would like to know what you think about.

Let $f,f_n:[0,1]\to \mathbb R$ be continuous functions. Assume that for every sequence $(x_n)_{n\in \mathbb N}$ converging to a point $x\in [0,1]$, we have $$\lim_n f_n(x_n)=f(x).$$ Can we say that $f_n$ converges uniformly to $f$ in $[0,1]$?

First of all, do you think the question is well-posed? I'm stuck in thinking about it and I don't manage to prove it neither to find a counterexample. Thanks for your help.

Answer 1

Yes, this is true:

Suppose that the convergence $f_n \to f$ is not uniform but that $f(x) = \lim_{n\to\infty}f_n(x_n)$ for every convergent sequence $x_n\to x$ in $[0,1]$.

After passing to a subsequence of the $f_n$ we may assume that there is $\varepsilon \gt 0$ such that for all $n \in \mathbb{N}$ we have $\sup_{x \in [0,1]} \lvert f(x) - f_n(x) \rvert \gt \varepsilon$. This implies that there is a sequence of points $x_n \in [0,1]$ such that $\lvert f(x_n) - f_n(x_n) \rvert \gt \varepsilon$ for each $n$. After passing to further subsequences, we may assume that $x_n \to x \in [0,1]$ by compactness of $[0,1]$. But then we're in trouble: our hypotheses tell us that $$\tag{$\ast$} \lim_{n\to \infty} \lvert f(x) - f_n(x_n)\rvert = 0, $$ which is incompatible with $f(x_n) \to f(x)$ and $\lvert f_n(x_n) - f(x)\rvert \gt \varepsilon$.

More formally: continuity of $f$ ensures together with $x_n \to x$ that there is $N$ such that $\lvert f(x) - f(x_n)\rvert \lt \varepsilon/2$ for all $n \geq N$ so that the triangle inequality gives us for all $n \geq N$ that $$ \begin{align*} \lvert f_n(x_n) - f(x) \rvert &= \lvert [f_n(x_n) - f(x_n)] - [f(x) - f(x_n)]\rvert \\ & \geq \underbrace{\lvert f_n(x_n) - f(x_n)\rvert}_{\gt \varepsilon} - \underbrace{\lvert f(x) - f(x_n)\rvert}_{\lt \varepsilon/2} \\& \gt \varepsilon/2 \gt 0, \end{align*} $$ which is absurd in view of $(\ast)$.

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Comments:

1. The mode of convergence $f_n(x_n) \to f(x)$ for all $x_n \to x$ is called continuous convergence. Continuity of the $f_n$ is never used in the above argument. In fact, $f_n(x_n) \to f(x)$ for all $x_n \to x$ ensures continuity of $f$ independently of the continuity of the $f_n$ (exercise).

2. You can find a generalization of the above result and a more detailed discussion in the section Compact and continuous convergence (Chapter 3, §1, section 5) starting on page 98 in R. Remmert's Theory of complex functions. The precise result reads that $f_n$ converges to $f$ continuously if and only if $f$ is continuous and $f_n \to f$ uniformly on compact sets. More details in loc. cit. (don't miss the historical notes preceding and following that section!).

Answer 2

There is a classical theorem by Dini that says this is possible when $f_{i}$ are monotonically increasing.

We now assume $f_{n}$ converges to $f$ pointwise and $f,f_{n}$ are all continuous but the limit is not uniform. Thus there is some $\epsilon$ such that $\forall N$ there is some $n\ge N$ such that $$|f_{n}(x)-f(x)|> \epsilon$$ for some $x\in [0,1]$.

Now since $f_{i},f$ are uniformly continuous on $[0,1]$, for $\epsilon_{1}=\frac{\epsilon}{3}$ we have a list of $\delta_{i}$s such that $0< \delta_{i}\le 1$ and $|f_{i}(x)-f_{i}(y)|\le \epsilon_{1}$ for $|x-y|\le \delta_{i}$. Consider the interval $U_{i}=[0,\delta_{i}]$, I claim $\lim_{i\rightarrow \infty} \bigcap U_{i}\not=\emptyset$. Suppose $\cap_{i=N} U_{i}=\emptyset$, then $f_{i}$ would have nontrivial variation $$\lim_{x \rightarrow y} |f(x)-f(y)|=0$$ a fact contradicting they are being continuous. Thus let $U=[0,\delta_{A}]$ be the intersection, and let $\delta_{B}$ be the corresponding value for $f$. We choose $\delta=\min[\delta_{A},\delta_{B}]$.

We now assume the counterexample in the beginning. For fixed $y$ such that $|x-y|\le \delta$ we can choose $N$ large enough such that $$|f(y)-f_{n}(y)|\le \epsilon/3$$ for any $n\ge N$. If it turns out $N$ is greater than the initial $n$ we chose, we may switch to a different $n$ guaranteed to exist by the hypothesis.

We have the estimate that $$|f_{n}(x)-f(x)|\le |f_{n}(x)-f_{n}(y)|+|f_{n}(y)-f(y)|+|f_{n}(y)-f_{n}(x)\le \epsilon$$ which contradicts the hypothesis we had earlier. So the convergence is uniform afterall.