Let $\left(f_{h}\right)_{h} \subset \mathcal{C}^{1}([a, b]) . $ Suppose we have by assumption the sets of functions $$ \left\{f_{h}: h \in \mathbb{N}\right\} \quad and \quad\left\{f_{h}^{\prime}: h \in \mathbb{N}\right\} $$ relatively compact in $\left(\mathcal{C}^{0}([a, b]),\|\cdot\|_{\infty}\right)$ Then $\exists\left(h_k\right)_{k}$ increasing sequence of integers and $\exists g_{1}, g_{2} \in \mathcal{C}^{0}([a, b])$ such that $$ f_{h_{k}} \longrightarrow g_{1} \text { for } k \rightarrow+\infty \text { in }\left(\mathcal{C}^{0}([a, b]),\|\cdot\|_{\infty}\right) $$ and $$ f_{h_{k}}^{\prime} \longrightarrow g_{2} \text { for } k \rightarrow+\infty \text { in }\left(\mathcal{C}^{0}([a, b]),\|\cdot\|_{\infty}\right) $$

What I have is that the $ f_{h_{k}}^{\prime}$ now are actually not the derivatives of the $ f_{h_{k}}$ and my goal is to be able to extract from the sequences $ f_{h_{k}}$ a subsequence such that the $ f_{h_{k}}^{\prime}$ of my new subsequence are actually their derivatives. My goal is to apply the Cantor diagonal procedure on a dense set of $[a,b]$, but I have difficulties in formalizing it. Thanks in advance!


1 Answer 1


No need for diagonal procedure. First pick a subsequence along which $(f_h)$ converges in the norm. Look at the derivatives along this subsequence. There is a further subsequence along which $(f_h')$ converges. You now have one subsequence along which both $(f_h)$ and $(f_h')$ converge.

Now use the fact that $f_{h_k} (x)=f_{h_k} (c)+\int_c^{x} f_{h_k}' (y)dy$ to conclude that $g_1$ is differentiable and its derivative is $g_2$.

  • $\begingroup$ I understand your point. So can I see it as just the first step of the diagonal process? In this sense, I would need a full diagonal process in the case I had to prove a similar theorem but for infinite differentiable functions (so that I get an infinite extraction)? $\endgroup$ Jan 17, 2021 at 10:08
  • 1
    $\begingroup$ @TheTurtleHermit True. For infinite differentiabiltiy you need diagona procedure. $\endgroup$ Jan 17, 2021 at 11:39

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