Completeness of ${C^2[0,1]}$ with under a specific metric 
Prove that ${C^2[0,1]} $ (set of two times differentiable functions)is complete with metric: $$d(f,g)=\sup_{x \in [0,1]}|f(x)-g(x)|+ \sup_{x \in [0,1]}|f'(x)-g'(x)| + \sup_{x \in [0,1]}|f''(x)-g''(x)|.$$

What I have trouble is taking a Cauchy sequence in the space and proving that it converges in that space. The second part of my question is proving that $$S=\{f(x)\in \mathbb{C^2[0,1]} \mid {\textstyle \int_{0}^{1}}f(x)dx\leq 3\} \subset {C^2[0,1]}$$ is complete under this metric. I am told to prove: 

If we have $\{f_n\}\subset S$ with $f_n \to f$ in ${C^2[0,1]}$, then $f \in S$.

Why is this true? Can anyone prove these?
 A: Hint.
Use two ingredients to prove the completness of $\mathcal C^2([0,1])$:


*

*Ingredient one the space $\mathcal C([0,1]$ of continuous real functions defined on $[0,1]$ is complete for the $\Vert f \Vert =\sup\limits_{x \in [0,1]} \vert f(x) \vert$ norm.

*Ingredient two if a sequence of differentiable functions $(f_n)$ is such that $f_n^\prime$ converges uniformly to $g$ and $f_n(x_0)$ converge for at least one point $x_0 \in [0,1]$ then $(f_n)$ converges uniformly to a function $f$ which is differentiable and $f^\prime=g$.


Some possible steps for the recipe.


*

*Take a Cauchy sequence $(f_n)$.

*By definition of your norm, $(f_n)$, $(f_n^\prime)$ and $(f_n^{\prime \prime})$ are Cauchy sequences.

*With ingredient 1, they converge uniformly to continuous functions $f$, $g$ and $h$ respectively.

*With ingredient 2, $g$ is differentiable and $g^\prime=h$.

*Again with ingredient 2 $f$ is differentiable and $f^\prime=g$.

*You can then conclude.


Regarding the second part of your question, prove that $S$ is closed.
