Show that $\phi(r) = ||f||_{L^r(0,1)}$, $r\in [1,2]$ is continuous function of r.

Question

Show that the function $$\phi(r) = ||f||_{L^r(0,1)}$$ for $r\in [1,2]$ is a continuous function of $r$.

Do I need to use the definition of continuity to solve this question? I don't know how to start this question.

Answer 1

We begin by assuming that the end point cases $\phi(1)$ and $\phi(2)$ are both finite. This guarantees $\phi(r)<\infty$ for all $1\le r\le 2$. Let $\Phi(r) = \phi(r)^r$. And note that $$F(x,y) = x^{\frac{1}{y}}=e^{\frac{ \log x}{y}} $$ is jointly continuous on $x\in [0,\infty), y\in (0,\infty)$. Since $\phi(r) = F(\Phi(r), r)$, it suffices to show $\Phi(r)$ is continuous. Note that it is sufficient to prove $\Phi(r_n)\to \Phi(r)$ whenever $r_n \to r$. This follows easily from dominated convergence theorem by noticing that $$ 0\le |f(x)|^r = |f(x)|^r 1_{\{|f(x)|<1\}}+|f(x)|^r 1_{\{|f(x)|\ge 1\}}\le |f(x)| 1_{\{|f(x)|<1\}}+|f(x)|^2 1_{\{|f(x)|\ge 1\}}\in L^1(0,1). $$ and $r \mapsto |f(x)|^r$ is continuous.

Note: However, we can say more about $\phi(r)$ actually. By Holder's inequality, we can show that $$ \|f\|_{{p_t}}\le \|f\|_{{p_0}}^{1-t}\|f\|_{{p_1}}^t $$ whenever $\frac{1}{p_t}=\frac{1-t}{p_0}+\frac{t}{p_1}$, $p_0, p_1 \in [1,\infty]$. This shows that $$ \psi:\frac{1}{p}\mapsto \log \|f\|_{p}, $$ is convex, and thus is continuous on$ \{\psi(x) <\infty\}$. Therefore, it follows that $\phi(r)=e^{\psi(\frac{1}{r})}$ is also continuous.