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

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

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

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

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

• Let $$K := \sup_n \sup_s \lvert fn'(s)\rvert\,.$$ Start by showing that $\lvert f(x) - f(y)\rvert \leqslant K\cdot \lvert x-y\rvert$. – Daniel Fischer May 31 at 15:03
• @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.... – codelearner Jun 1 at 2:38

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.
• 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. – Saptak Bhattacharya Jun 1 at 3:45