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.

  • $\begingroup$ Is this a Riemann-Stieltjes integral? $\endgroup$ – xbh Aug 13 '18 at 4:07
  • $\begingroup$ Use squeeze theorem. $\endgroup$ – xbh Aug 13 '18 at 4:08
  • $\begingroup$ @xbh, actually, Riemann-Stieltjes not part of the content, but, by the "appearance" of the integral, it seems that it's really (this question is of old exam, perhaps it has been recently withdrawn from the content). But, would it be possible to solve without using the generalization of integral? $\endgroup$ – Corrêa Aug 13 '18 at 4:15
  • $\begingroup$ @xbh My ideia is to use the sequeeze theorem, but I couldn't find $f_ {n}$ and $g_{n}$ $\endgroup$ – Corrêa Aug 13 '18 at 4:17

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 $M^2$. $\blacktriangleright$

  • $\begingroup$ @RRL Sorry about this. I took too long on typing, and I didn't see your hint at first. My bad. $\endgroup$ – xbh Aug 13 '18 at 4:43
  • $\begingroup$ Not bad at all. IMO, independent answers are always acceptable. $\endgroup$ – marty cohen Aug 13 '18 at 4:47
  • $\begingroup$ That's understandable. Its a good answer so I'll give it an upvote, but I not inclined to fully answer such a question. $\endgroup$ – RRL Aug 13 '18 at 4:47
  • $\begingroup$ @RRL If you thought i post this after i saw your hint, then you could post your hint and i could delete mine. I do not mind. $\endgroup$ – xbh Aug 13 '18 at 4:50
  • $\begingroup$ @xbh: Not a problem. $\endgroup$ – RRL Aug 13 '18 at 4:52

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