First question here so really excited and hope you can help me, thanks!
In my intro to functional analysis class we just now covered $L^p$ spaces and I was presented with this homework question:
Given $ 0 < p < 1 $, show the formula $$ \rho(f,g)= \int |f-g|^p\mathsf dx $$ defines a metric and the resulting topological vector space is complete.
I don't think I quite got $L^p$ spaces yet and so I really hope there is someone who can take it easy on me and help me with this. Thanks all
EDIT: I should have mentioned triangle inequality and completeness of the space are where I am stuck
Given a measure space $(X,\mathcal M,\mu)$, we define $$L^1(X) = \left\{f : X\to\mathbb C \;\mid\; \int_X |f|\mathsf d\mu < \infty\right\}$$ and $$\mathcal L^1(X) = L^1(X) / \sim $$ where $\sim$ is the equivalence relation on the bracketed set defined by $f\sim g$ iff $f=g$ a.e. That is, $$\mu(\{x\in X : f(x)\ne g(x)\}) = 0. $$ In other words, we identify two integrable functions $f$ and $g$ if they are equal almost everywhere. Generally this distinction isn't made explicit, and $L^1$ is used to denote both of these (since it is cumbersome to refer to equivalence classes instead of functions). Given $0<p<\infty$, define $$L^p(X) = \left\{f:X\to\mathbb C \mid f^p \in L^1(X)\right\}. $$ That is, $f\in L^p(X)$ iff $$\int_X |f|^p \mathsf d\mu < \infty,$$ and define $$\mathcal L^p(X) = L^p(X)/\sim. $$
Suppose now $0<p<1$. Since a nonnegative measurable function $f$ has zero integral iff $f=0$ a.e. it is clear that for $f,g\in\mathcal L^p(X)$ $$\int_X |f-g|^p\mathsf d\mu = 0 \iff f=g, $$ where we interpret $f=g$ as $f=g$ a.e. (note that without the technical point about equivalence classes, we would only have a pseudometric space). The second condition of a metric space is clear from $|f-g|^p = |g-f|^p$. If $f,g,h\in\mathcal L^p(X)$ then \begin{align} \rho(f,h) &= \int_X |f-h|^p \mathsf d\mu\\ &= \int_X |f-g+g-h|^p \mathsf d\mu\\ &\leqslant \int_X (|f-g|^p + |g-h|^p)\mathsf d\mu \tag1\\ &= \int_X |f-g|^p\mathsf d\mu + \int_X |g-h|^p\mathsf d\mu, \end{align} where $(1)$ is justified by the inequality $(a+b)^p \leqslant a^p + b^p$ for $a,b\geqslant0$.
Now, $\rho$ induces a topology on $X$ with the basis consisting of all open balls, i.e. sets of the form $$B(f;\varepsilon) := \{g\in X : \rho(f,g) <\varepsilon\}. $$
Edit: I read the question as "show that $X$ is a topological vector space," as opposed to "show that the resulting topological vector space is complete." So to address that notion:
We need to show that if $\{f_n\}$ is a Cauchy sequence in $\mathcal L^p(X)$ then there exists $f\in\mathcal L^p(X)$ with $\lim_{n\to\infty} f_n = f$. Since $\{f_n\}$ is a Cauchy sequence, we can construct a subsequence $\{f_{n_k}\}$ such that $\rho(f_{n_k}, f_{n_{k+1}}) \leqslant 2^{-k}$ for $k\in\mathbb N$. Define $$ f = f_{n_1} + \sum_{k=1}^\infty (f_{n_{k+1}}-f_{n_k}). $$ Then for any $N\in\mathbb N$, $$\rho\left(\sum_{k=1}^N f_{n_{k+1}},\; \sum_{k=1}^N f_{n_k}\right)\leqslant \sum_{k=1}^N \rho(f_{n_{k+1}},f_{n_k})\leqslant \sum_{k=1}^N 2^{-k}\leqslant 1, $$ by the triangle inequality. Put $$g_N= |f_{n_1}|+\sum_{k=1}^N |f_{n_{k+1}}-f_{n_k}|.$$ Then $g_N\leqslant g_{N+1}$ and $$g:=\lim_{N\to\infty} g_N = |f_{n_1}|+\sum_{k=1}^\infty |f_{n_{k+1}}-f_{n_k}|, $$ so by monotone convergence \begin{align} \int_X \left(|f_{n_1}|+\sum_{k=1}^\infty |f_{n_{k+1}}-f_{n_k}|\right)^p \mathsf d\mu &= \lim_{N\to\infty}\int_X \left(|f_{n_1}|+\sum_{k=1}^N |f_{n_{k+1}}-f_{n_k}|\right)^p\mathsf d\mu\\ &\leqslant \int_X |f_{n_1}|^p\mathsf d\mu + \lim_{N\to\infty}\int_X \left(\sum_{k=1}^N |f_{n_{k+1}}-f_{n_k}|\right)^p\mathsf d\mu\\ &\leqslant \int_X|f_{n_1}|^p\mathsf d\mu + \lim_{N\to\infty}\sum_{k=1}^N \int_X |f_{n_{k+1}}-f_{n_k}|^p\mathsf d\mu\\ &= \int_X|f_{n_1}|^p\mathsf d\mu + \sum_{k=1}^\infty \rho(f_{n_{k+1}},f_{n_k})\\ &\leqslant \int_X|f_{n_1}|^p\mathsf d\mu + 1 < \infty, \end{align} so that $g = \lim_{N\to\infty}g_N \in\mathcal L^p(X)$. Now $$|f| = \left| f_{n_1} + \sum_{k=1}^\infty (f_{n_{k+1}}-f_{n_k})\right| \leqslant |f_{n_1}| + \sum_{k=1}^\infty |f_{n_{k+1}}-f_{n_k}| = g, $$ so that $f\in\mathcal L^p(X)$. Finally, as $\{f_n\}$ is a Cauchy sequence with a subsequence that converges to $f$, it follows that $\lim_{n\to\infty} f_n=f$.