Continuity and $L^p$ spaces

Question

I have been wondering how to solve this question I saw in a textbook. Given $ g \in \bigcup _{1\leq p\leq \infty} L^{p}$ define, for $ r \in [ 0,1]$ , $$ G(r) = \int_{0}^{r} g(t) dt \;.$$ Show that $G$ is continuous on $ [0,1]$ and that if $ g\in L^{p}$ for some $ p> 1$ , then $G \in \lambda_{{1/p'}} [0, 1]$.

Answer 1

Since $[0,1]$ has finite measure, $g \in \bigcup L^p[0,1]$ implies that $g \in L^1[0,1]$. (If you meant $g \in L^p(\mathbb{R})$ for some $p$ notice that its restriction to $[0,1]$ is still in $L^1[0,1]$). Define $G_x = g \cdot \chi_{[0, x]}$ where $\chi_{[0,x]}$ is the characteristic function of $[0,x]$. Then $$G(r) = \int_0^1 G_r (t) dt$$ Choose some convergent sequence $x_n \to x$ in $[0,1]$. Then $G_{x_n} \to G_x$ pointwise and $G_{x_n}$ is dominated by $g$. Then by the dominated convergence theorem we have $$G(x_n) = \int_0^1 G_{x_n}(t) dt \to \int_0^1 G_x(t) dt = G(x)$$ which proves that $G$ is continuous. For the Hölder continuity, notice that $$|G(x) - G(y)| = \left|\int_0^x g(t) dt - \int_0^y g(t) dt\right| = \left|\int_y^x g(t) dt\right|$$ $$ = \left|\int_y^x 1 \cdot g(t) dt \right| \le \|g\|_{L^p[y,x]}\|1\|_{L^{p'}[y,x]}$$ $$ \leq \|g\|_p|x-y|^{1/p'}$$ by Hölder's inequality. Here we have implicitly assumed $p > 1$, since for $p=1$ it is not true that $\|1\|_{L^{p'}[y,x]} = |x-y|^{1/p'}$ as in this case $p' = \infty$.