Let $a$ and $b$ be given real numbers such that $a < b$; let $p$ be a given positive real number; let $X$ denote the set of all the (real or) complex-valued functions that are defined and continuous on the closed interval $[a, b]$ on the real line; and let the function $d \colon X \times X \to \mathbb{R}$ be defined as $$ d(x, y) \colon= \sqrt[p]{ \int_a^b \lvert x(t) - y(t) \rvert^p \ \mathrm{d} t } $$ for all $x, y \in X$.
Then for which values of $p > 0$ is this function $d$ a metric? I know that this function is indeed a metric for $p \geq 1$. What about the values of $p \in (0, 1)$?
And, for what values of $p$ for which the function $d$ is a metric is the metric space $(X, d)$ a complete metric space? I know that for $p=1$, $a= 0$, and $b = 1$, we have a non-complete metric space, as is proved in Example 1.5-9 in the book Introductory Functional Analysis With Applications by Erwine Kreyszig. How to proceed in the general case? My feeling is that this space is non-complete for all $p \geq 1$, but I'm unable to show this (rigorously enough).
I would appreciate an answer in as much rigor and detail as possible.
Here is the link to my Mathematics Stack Exchange post where the special case when $p = 1$ has been addressed.
Is this metric space complete?
From this post, I know that our metric space is not a complete metric space for $p = 1$, for any real numbers $a$ and $b$ such that $a < b$, but nobody over there has actually shown the details of the proof.
The expression $$d(x, y) \colon= \sqrt[p]{ \int_a^b \lvert x(t) - y(t) \rvert^p \ \mathrm{d} t }$$ is not a metric when $0<p<1$. Indeed, let $a=0$ and $b=1$ for simplicity and choose $x = \chi_{[0,1/2]}$ and $y=\chi_{[1/2,1]}$. Then $d(0,x) =d(0,y)= 1/2^{1/p}$ while $d(x,y)=1$. Since $p<1$, $$d(x,0)+d(y,0) = 2^{1-1/p} < 1 = d(x,y)$$ violating the triangle inequality.
However, the quantity $$\rho (x, y) \colon= \int_a^b \lvert x(t) - y(t) \rvert^p \ \mathrm{d} t $$ is a metric for $0<p<1$. Indeed, the function $g(s) = s^p$ is increasing and concave for positive arguments $s$, which implies it is subadditive: $g(s_1+s_2) \le g(s_1)+g(s_2)$. Hence, for any three functions $x,y,z$ we have $$ \lvert x(t) - y(t) \rvert^p \le \lvert x(t) - z(t) \rvert^p + \lvert z(t) - y(t) \rvert^p $$ pointwise, and then integration proves the claim.
Completeness
Using any kind of integral norm on the space of continuous functions results in an incomplete space, because convergence with respect to integral norm allows sharp peaks to form, destroying continuity. A typical example is $x_n(t)=\min(t^n,1)$ on $[0,2]$, which is Cauchy with respect to any $L^p$ norm for $1\le p<\infty$ and also with respect to the metric $\rho$ for $0<p<1$. It does not converge in the space $C[0,2]$, since its limit would be $\chi_{[1,2]}$, discontinuous.
The case $p=\infty$ is an exception: uniform norm plays well with continuity. We just get $C[a,b]$ with uniform norm, a complete space.
To get completeness for all $p$, one needs to allow Lebesgue measurable functions with finite integral of $|x|^p$. Then all of the metric spaces considered above are complete.