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Assume that $(f_n)_n$ is a sequence of functions continuous on $[0$, $1]$, differentiable on $(0,1)$, that converge pointwise on that interval to a function $f$, and such that each $f_n$′ is bounded on $(0$, $1)$ and the sequence $(sup_{s\in(0,1)} |f_n′ (s)|)_n$ is bounded.

Setting $g_n = f_n − f$ , we know that $g_n$ → 0 pointwise on $[0$, $1]$.

Show that there exists a real number $R≥0$ such that, for all $x$, $y\in[0,1]$ and all $n \in \mathbb{N}$, $|g_n(x)−g_n(y)|\le R|x−y|$.

My attempt

Here I used the Mean Value theroem as $g_n$ is continuous and differentiable because of $f_n$. So for $x$, $y \in [0,1]$, $g_n'(c)$ = $|g_n(x) - g_n(y)| \over |x -y| $. This can be rewritten as $g_n'(c) |x - y|$ = $|f_n(x) − f(x)| - |f_n(y) − f(y)|$. By pointwise convergence and as n -> infinity then $g_n'(c) |x - y|$ = $|f(x) − f(x)| - |f(y) − f(y)|$ which means $g_n'(c) |x - y|$ = 0 thus, $g_n'(c)$ is either $0$ or positive hence $g_n'(c) = R$.

Edit: I am rethinking my answer because I can't prove $g_n$ is continuous as I don't know if the limit function $f$ is continuous or not. Any push in the right direction is greatly appreciated.

Edit 2: Thank you for the hint by @Saptak Bhattacharya and @Daniel Fischer. I tried to use the triangle of inequality but I am not sure if what I did was right. This is how I did it:

$|g_n(x) - g_n(y)| = |f_n(x) - f(x) - (f_n(y) - f(y))|$ $\leq$ $|f_n(x) - f_n(y)| + |f(y) - f(x)| $

$|f_n(x) - f_n(y)|$ is bounded by $M |x -y|$ (I've shown this by previous question).

Hence $|f_n(x) - f(x) - (f_n(y) - f(y))|$ $\leq$ $M |x -y| + |f(y) - f(x)| $.

$|f_n(x) - f(x) - (f_n(y) - f(y))|$ $\leq$ $M |x -y| + |f(y)| + |f(x)| $.

$|f_n(x) - f_n(y)| + |f(x)| + |f(y)|$ $\leq$ $M |x -y| + |f(y)| + |f(x)| $.

$|f_n(x) - f_n(y)| \leq M |x -y|$

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  • $\begingroup$ Let $$K := \sup_n \sup_s \lvert fn'(s)\rvert\,.$$ Start by showing that $\lvert f(x) - f(y)\rvert \leqslant K\cdot \lvert x-y\rvert$. $\endgroup$ – Daniel Fischer May 31 at 15:03
  • $\begingroup$ @DanielFischer Thank you for the hint. So I did show 𝐾⋅|𝑥−𝑦| is an upper bound by a previous question and I have edited my answer to show the same for $g_n$ using triangle of inequality but I am not sure if I did it right.... $\endgroup$ – codelearner Jun 1 at 2:38
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Oh $f$ is indeed continuous,infact,the pointwise limit of a sequence of uniformly Lipschitz functions is always continuous.To see why,first observe,what did actually prevent the preservation of continuity in pointwise limits?Indeed, you made the functions peak sharply at a point to a fixed value, didn't you?Observe that in such a construction,you had no control over the rate of change of $f_n$ at that particular point as $n\to \infty$.This suggests, that somehow if we can control the rates of change of $f_n$ in the limit,we can have a way of preserving continuity under pointwise limits.Infact, the limit $f$ shall itself be Lipschitz.The easiest way to observe this is observing that $|f_n(x)-f_n(y)|\leq M|x-y|$ for a fixed $M>0$(in your case, $M$ happens to be the supremum of the absolute values of the derivatives), all $x$ and $y$, and natural numbers $n$, and then taking limit as $n\to \infty$.Since we have got that $f$ is Lipschitz too, we can prove that $g_n$ is uniformly Lipschitz by a simple application of the triangle inequality.

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  • $\begingroup$ I tried using triangle of inequality but I am not sure if what I did was right so I ended up not using it. After your hint, I went back and tried using it again. I am not sure if what I did was right.... $\endgroup$ – codelearner Jun 1 at 2:10
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    $\begingroup$ Since we have already shown that $f$ is Lipschitz,with Lipschitz constant $M$ and $f_n$ is uniformly Lipschitz with the same Lipschitz bound,we apply triangle inequality to get $|g_n(x) - g_n(y)|=|f_n(x)-f_n(y)+f(y)-f(x)|\leq |f_n(x)-f_n(y)|+|f(x)-f(y)|\leq 2M|x-y|$.Did you do this?If yes,then it's alright. $\endgroup$ – Saptak Bhattacharya Jun 1 at 3:45

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