I'm having trouble determining if $X$ is complete WRT $||.||_{\infty}$ norm. I know that in order to show that I need to take a Cauchy sequence and show that it has a limit in my space $X$ or find a Cauchy sequence of functions which satisfy $f(0)=f'(0)$ and which limit lies outside of $X$. I tried but was unable to find a counter example, therefore I believe the statement to be true, however I am not sure exactly how to go about proving it.
So far I was able to prove that any $g\in X$ is of the form $f(0)+\int_0^xf(y)dy$ where $f\in C[0,1]$ if it helps.
would really appreciate it if somebody could show me how to prove of disprove this statement. Thanks
No, it is not complete.
For each $n\in\Bbb N$, define$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}nx^2&\text{ if }x<\frac1{2n}\\x-\frac1{4n}&\text{ otherwise.}\end{cases}\end{array}$$Clearly, each $f_n$ belongs to $X$. But, in $\bigl(C^1[0,1],\lVert\cdot\rVert_\infty\bigr)$, the sequence $(f_n)_{n\in\Bbb N}$ converges to $f(x)=x$. And $f\notin X$. So, $X$ is not a closed subspace of $C^1[0,1]$ and therefore it is not complete. Or you can say that $(f_n)_{n\in\Bbb N}$ is a Cauchy sequence of elements of $X$ which does not converge in $X$.