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, 2018 at 4:07
  • $\begingroup$ Use squeeze theorem. $\endgroup$
    – xbh
    Aug 13, 2018 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the "appearance" of the integral, it seems that it's really (this question is of old exam, perhaps it has been recently withdrawn from the content).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? $\endgroup$
    – Lucas
    Aug 13, 2018 at 4:15
  • $\begingroup$ @xbh My ideia is to use the sequeeze theorem, but I couldn't find $f_ {n}$ and $g_{n}$ $\endgroup$
    – Lucas
    Aug 13, 2018 at 4:17

1 Answer 1


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, 2018 at 4:43
  • $\begingroup$ Not bad at all. IMO, independent answers are always acceptable. $\endgroup$ Aug 13, 2018 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, 2018 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, 2018 at 4:50
  • $\begingroup$ @xbh: Not a problem. $\endgroup$
    – RRL
    Aug 13, 2018 at 4:52

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