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Let $\{f_k\}_{k\ge 0}$ be a sequence of continuous and bounded functions $f_k\colon \mathbb{R}\to \mathbb{R},\ x\mapsto f_k(x)$, converging pointwise to $f$. Suppose that there exists a sequence of increasing integers $k_1,k_2,\dots$ and a corresponding convergent sequence $x_1,x_2,\dots$ such that $\displaystyle\lim_{\ell\to\infty}x_\ell = x$ and $\displaystyle\lim_{\ell\to\infty} f_{k_\ell}(x_\ell)$ exists.

My question: Is it true that $$ \lim_{\ell\to\infty} f_{k_\ell}(x_\ell) = f(x)\ \ \ ? $$


I apologize if this is a silly question. Any help is very appreciated.

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No, consider f.e. the functions $f_n(x)=x^n$ on $[0,1]$, and $x_n=1-1/n$. It follows that $x_n \to x=1$ and $f_n(x_n)=(1-1/n)^n \to 1/e$, but $1^n\to 1$. To ensure that the limit of a converging sequence of continuous functions is itself continuous, you need uniform convergence.

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  • $\begingroup$ I see, thanks for your answer! Just a side comment: What happens if the pointwise limit $f$ is a continuous function? $\endgroup$ – Ludwig Jul 24 '17 at 7:44
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If each $f_k$ is defined on a closed and bounded interval and $(f_k)$ is uniformly bounded and equicontinuous, then by the Arzela-Ascoli theorem there exists a uniformly convergent subsequence. Under these very restricted conditions the result is true for that subsequence.

There are counterexamples where the sequence converges to a continuous function. For example, take $f_k(x) = k|x| e^{-k|x|}$. We have pointwise convergence where $f_k(x) \to f(x) = 0.$ However, for any subsequence $(f_{k_l})$ taking $x_l = 1/n_{k_l}$, we have $x_l \to 0$ but

$$\lim_{l \to \infty}f_{k_l}(x_l)= e^{-1} \neq f(0) = 0$$

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Assuming the limit is continuous changes nothing.

Consider the "$n$th hat" $\Lambda_n:[0,1]→[0,1]$ whose $(x,y)$-graph, drawn not to scale, $$\wedge\text{_} $$ is made up of 3 straight lines, the first joining $(0,0)$ to $(1/n,1)$ and the second joining $(1/n,1)$ to $(2/n,0)$, and then 0 for $x>2/n$. In symbols, $$ \Lambda_n(x) = \begin{cases} nx & x∈ [0,1/n] \\ 1-n(x-1/n) & x-1/n∈[0,1/n] \\ 0 & x-2/n > 0 \end{cases} $$ Then observe $\Lambda_n(1/n) = 1 \xrightarrow[n→∞]{} 1$, yet $\Lambda_n → 0$ pointwise.

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