Let $X=C^{1}[0, 1]$. For each of $f\in X$, define $$p_1(f):= \sup\{{|f(t)|:t\in [0, 1]}\}$$ $$p_2(f):= \sup\{{|f'(t)|:t\in [0, 1]}\}$$ $$p_3(f):=p_1(f)+p_2(f)$$ I know that $(X, p_1 ), (X, p_2 )$ are not Banach spaces. Can I claim that $p_3(f)$ is also not a Banach space since each one is not a Banach space? Please help me as I am self studying functional analysis.
Thank you very much for your time.
Under $p_3$, $X$ is a Banach space: let $\{f_n\}$ be a Cauchy sequence. Note that $\{f_n\}$ is Cauchy under $p_1$ and $p_2$. By completeness of $C[0,1]$, the sequence $\{f_n\}$ converges uniformly to some continuous function $f$. Similarly, $\{f'_n\}$ converges uniformly to some continuous function $g$. Now, by letting $n \to \infty$ in the equation $f_n(x)=f_n(0)+\int_0^{x} f'_n(t)\, dt$ we get $f(x)=f(0)+\int_0^{x} g(t) \, dt$. This implies that $f$ is differentiable and $g=f'$. It now follows from definition that $p_3(f_n -f) \to 0$.