How to obtain sequences in $C^1[a,b]$ which converges to some functions in $C[a,b]$ with some conditions Given $\chi$ and $\chi_1$ in $C[a,b]=\{\phi:[a,b]\to\mathbb{R} : \phi \text{ is continuous}\}$ satisfying $\chi(t)=\chi_1(t), \forall t\in[a,c]$, where $c\in (a,b)$. How i can get two sequences $(\chi^m),(\chi_1^m)\subset C^1[a,b]=\{\phi:[a,b]\to\mathbb{R} : \phi \text{ is continuously differentiable}\}$ with $\chi^m(t)=\chi_1^m(t),\forall t\in [a,c]$ which converges (in sup norm) to $\chi$ and $\chi_1$ respectively.
My attempt: Using Stone–Weierstrass theorem, we can obtain a sequence ${\chi}^{m}\in C[a,b]$ which converges to $\chi$. Now the problem is how to define $\chi_1^m$ in $(c,d]$ in order to obtain $\chi_1^m \to \chi_1$.
 A: The function $f:[a,b]\to \mathbb{R}^n$ consists of the family of $n$ coordinate functions $f_j:[a,b]\to \mathbb{R},$ so it suffices to consider every function separately, i.e. the problem can be reduced to $n=1.$
We will make use of the following.
Lemma Given $\delta>0$ and $t\in \mathbb{R}.$ There exists a   function $u\in C[a,b],$  such that $u'$ is continuous on the intervals $[a,c],$ $[c,b]$ and $$u'_+(c)= t,\quad u(x)=0, \ a\le x\le  c,\quad |u(x)|\le \delta, \ a\le x \le b.$$
Indeed, let $$u(x)=\begin{cases} 0 & a\le x\le c \\ {t\over n} [1-e^{-n(x-c)}]& c<x\le b\end{cases} ,$$
Then $u(x)$ is continuous and $$u'_+(c)= t,\qquad |u(x)|\le {|t|\over n}.$$ Choosing $n>2|t|/\delta $ gives the conclusion of Lemma.
We will apply the Bernstein polynomials for the intervals $[a,c]$ and $[c,b].$
For functions $f:[0,1]\to \mathbb{R}$ they are
defined
by $$B_n(f)=\sum_{k=0}^n f\left ({k\over n}\right ){n\choose k}x^k(1-x)^{n-k}.$$
It is known that $B_n(f)\rightrightarrows f.$ Moreover, we have $B_n(f)(0)=f(0)$ and $B_n(f)(1)=f(1).$
By the affine change of variables the Bernstein polynomials can be constructed for functions defined on any closed interval, in particular for $f\in C[a,c]$ and $f\in C[c,b].$
Consider the Bernstein polynomials
$B^{[a,c]}_n(f)(x)$ and $B^{[c,b]}_n(f)(x).$ According to the corresponding property on $[0,1],$ they satisfy $$B^{[a,c]}_n(f)(c)=B^{[c,b]}_n(f)(c)=f(c).$$
Therefore the function
$$g_n(x)= \begin{cases} B^{[a,c]}_n(f)(x) & a\le x\le c \\
B^{[c,b]}_n(f)(x) & c<x\le b
 \end{cases}
$$
are continuous and $$g_n(x)\rightrightarrows f(x),\quad a\le x\le b.$$
On each interval $[a,c]$ and $[c,b]$ the functions $g_n'$ are continuous (as polynomials).
We are going to modify $g_n(x)$ on the interval $[c,b]$ in order to gain two sided continuous differentiability  at $x=c.$
For a given $\delta>0$ choose $n$ such that $|f(x)-g_n(x)|< \delta .$ Next choose $u(x)$ satisfying
Lemma with $u'_+(c)= (g_n)'_-(c)-(g_n)'_+(c)$ and $|u(x)|\le \delta$ for $a\le x\le b.$ Then the function $h(x)=g_n(x)+u(x)$ belongs to $ C'[a,b],$ as
$$ h'_-(c)=(g_n)'_-(c)=h'_+(c).$$ Moreover
$$|f(x)-h(x)|\le |f(x)-g_n(x)|+|u(x)|<2\delta.$$
