Problem. Let $\phi: [a,b] \to \mathbb{R}$ continous and $\alpha:[a,b] \to \mathbb{R}$ strictly increasing. Show that $$\lim_{n \to \infty}\left(\int_{a}^{b}\phi^{2n}\mathrm{d} \alpha \right)^{\frac{1}{n}} = \sup_{a \leq x \leq b} \phi^{2}(x).$$
I never solved such a question, it seems very confusing to me. I know that $$\Vert f \Vert_{p} = \left(\int_{a}^{b}|f(x)|^{p}\mathrm{d}x\right)^{\frac{1}{p}}$$ defines norm in $V$ (where $V$ is a space of continuous functions). So, we can write $$\left(\int_{a}^{b}\phi^{2n}\mathrm{d} \alpha \right)^{\frac{1}{n}} = \left(\int_{a}^{b}|\phi^{2}|^{n}\mathrm{d} \alpha \right)^{\frac{1}{n}} = \Vert \phi^{2} \Vert_{n}.$$ Thus, the problem is reduced to $$\lim_{n \to \infty}\Vert \phi^{2} \Vert_{n} = \Vert \phi^{2} \Vert_{\infty} \tag{*}\label{*}$$ where $\Vert \phi^{2} \Vert_{\infty} = \{\phi^{2}(x) : x \in [a,b]\}$. Now
$$\Vert \phi^{2} \Vert_{n}^{n} = \int_{a}^{b}|\phi^{2}|^{n}\mathrm{d} \alpha \leq \int_{a}^{b}\Vert \phi^{2} \Vert_{\infty}^{n}\mathrm{d}\alpha = \Vert \phi^{2} \Vert_{\infty}^{n}\int_{a}^{b}\mathrm{d}\alpha = \cdots$$
and $\Vert \phi^{2} \Vert_{\infty} = \phi^{2}(x_{0})$, then
$$\Vert \phi^{2} \Vert_{\infty}^{n} = \phi^{2}(x_{0})^{n} \leq \cdots$$
My ideia is to show that $$f_{n}\Vert \phi^{2} \Vert_{\infty} \leq \Vert \phi^{2} \Vert_{n} \leq g_{n}\Vert \phi^{2} \Vert_{\infty}$$ where $f_{n}, g_{n} \to 1$ when $n \to \infty$, but I couldn't find these functions (to solve "$\cdots$").
I proved (*) for norm in $\mathbb{R}^{n}$, but it was simpler. It may be that I'm completely wrong, anyway, I would like help.
Proof. $\blacktriangleleft$ By the continuity of $\varphi$ on $[a,b]$, the supremum can be attained in the interval. Denote the maximum i.e. supremum by $M$. Then $$ \left(\int_a^b \varphi^{2n} \mathrm d \alpha\right)^{1/n} \leqslant \left(\int_a^b M^{2n} \mathrm d \alpha \right)^{1/n} = M^2 \left(\int_a^b \mathrm d \alpha \right)^{1/n} = M^2 (\alpha(b) - \alpha(a))^{1/n} \to M^2 [n \to \infty]. $$
Since $M$ could be attained on $[a,b]$, by continuity for all $\varepsilon > 0$ there is a subinterval $[c,d] \subseteq [a,b]$ s.t. $\varphi(x) > M - \varepsilon$ on $[c,d]$. Therefore, $$ \left(\int_a^b \varphi^{2n} \mathrm d \alpha \right)^{1/n} \geqslant \left(\int_c^d \varphi^{2n} \mathrm d \alpha \right)^{1/n} \geqslant (M - \varepsilon)^2 (\alpha(b) - \alpha(a))^{1/n} \to (M - \varepsilon)^2 [n \to \infty]. $$ Therefore, $$ \forall \varepsilon >0, (M - \varepsilon)^2 \leqslant {\underline \lim} \left( \int_a^b \varphi^{2n} \mathrm d \alpha\right)^{1/n} \leqslant \overline {\lim} \left( \int _a^b \varphi^{2n} \mathrm d \alpha\right)^{1/n} \leqslant M^2. $$ Therefore the limit exists and equals $M^2$. $\blacktriangleright$