The post contains a problem and my solution to it. I am confusing with choosing $c$ since it depends on $p$ and I chose $p$ such that $|c| < \frac{\epsilon}{2(b-a)}$. Could you please check my solution and correct my mistakes if there is any? Thank you so much.
For a bounded interval $[a,b] \subset \mathbb{R}$ consider a Lipschitz continuous function $f:[a,b] \to \mathbb{R}$. Show that for arbitrary $\epsilon >0$ there is a polynomial $g(x)$ satisfying $g(x) = f(a)$, $g(b) =f(b)$ and
$$\| g'\|_{L^\infty(a,b)} \le \| f'\|_{L^\infty}(a,b) + \epsilon, \quad \|f-g\|_{L^\infty(a,b)} + \| f' -g'\|_{L^1 (a,b)} < \epsilon$$
My solution It is clear that $f'$ is measurable function. By Lusin's theorem; given $\epsilon >0$ then there is a function $h:[a,b] \to \mathbb{R}$ such that
1- $h$ is continuous on $[a,b]$.
2- The measure of $\{x: h(x) \not= f'(x) \} $ is less than $ \epsilon$
3- $\| h\|_{L^\infty(a,b)} \leq \|f'\|_{L^\infty(a,b)}.$
By Weierstrass’ theorem there is a polynomial $p$ such that
$$\| h -p\|_{L^\infty(a,b)} < \frac{ \epsilon} {2}, \quad \,\,\text{ and }\,\, |c| < \frac{\epsilon}{2(b-a)}.$$
where $$c= \frac{f(b) - f(a)}{b-a}-\frac{1}{b-a} \int_a^b p(t)dt .$$
Now we define the function
$$g(x) := c (x-a) +f(a)+ \int_a^x p(t) dt, \quad c \in \mathbb{R}.$$
We have $g(a)= f(a)$, and $g(b) =f(b)$ and; \begin{align*} \| g'\|_{L^\infty(a,b)} &= \| c + p \|_{L^\infty(a,b)}\\ & =\| c + p -h + h\|_{L^\infty(a,b)}\\ & \leq \|c\|_{L^\infty(a,b)} + \| p-h\|_{L^\infty(a,b)} +\|h\|_{L^\infty(a,b)}\\ & \leq \frac \epsilon 2 + \frac \epsilon 2 + \|h\|_{L^\infty(a,b)}\\ &\leq \|f'\|_{L^\infty(a,b)} + \epsilon \end{align*}
\begin{align*} \lVert f-g \rVert_{L^\infty(a,b)} & = \lVert f - c(x-a) - f(a) - \int_a^x p(t) dt \rVert_{L^\infty(a,b)} \\ &= \lVert f(a) + \int_a^x f'(t) dt - c(x-a) +f(a) + \int_a^x p(t) dt \rVert_{L^\infty(a,b)}\\ & \leq (b-a) \lVert f' - p \rVert_{L^\infty(a,b)}\\ & \le (b-a) \left(\lVert f' -h\rVert_{L^\infty(a,b)} + \lVert h -p \lVert_{L^\infty(a,b)} \right)\\ &\leq \frac{\epsilon}{2}. \end{align*}
\begin{align*} \lVert f' -g' \rVert_{L^1(a,b)} &= \lVert f' -c - p \rVert_{L^1(a,b)}\\ & \le \lVert f' -p \rVert_{L^1(a,b)}\\ &\leq \lVert f'-h \rVert_{L^1(a,b)} + \lVert h -p \rVert_{L^1(a,b)}\\ &\leq (b-a) \left( \lVert f'-h \rVert_{L^\infty(a,b)} + \lVert h -p \rVert_{L^\infty(a,b)} \right)\\ & \leq \epsilon \end{align*}
Therefore, we get
$$ \lVert f-g \rVert_{L^\infty(a,b)} + \lVert f' -g' \rVert_{L^1(a,b)} \leq \epsilon.$$