Vtyjkyuk

Problem. Let $phi: [a,b] to mathbbR$ continous and $alpha:[a,b] to mathbbR$ strictly increasing. Show that
$$lim_n to inftyleft(int_a^bphi^2nmathrmd alpha right)^frac1n = sup_a leq x leq b phi^2(x).$$

I never solved such a question, it seems very confusing to me. I know that
$$Vert f Vert_p = left(int_a^b|f(x)|^pmathrmdxright)^frac1p$$
defines norm in $V$ (where $V$ is a space of continuous functions). So, we can write
$$left(int_a^bphi^2nmathrmd alpha right)^frac1n = left(int_a^b|phi^2|^nmathrmd alpha right)^frac1n = Vert phi^2 Vert_n.$$
Thus, the problem is reduced to
$$lim_n to inftyVert phi^2 Vert_n = Vert phi^2 Vert_infty tag*label*$$
where $Vert phi^2 Vert_infty = phi^2(x) : x in [a,b]$.
Now

$$Vert phi^2 Vert_n^n = int_a^b|phi^2|^nmathrmd alpha leq int_a^bVert phi^2 Vert_infty^nmathrmdalpha = Vert phi^2 Vert_infty^nint_a^bmathrmdalpha = cdots$$

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
$$f_nVert phi^2 Vert_infty leq Vert phi^2 Vert_n leq g_nVert phi^2 Vert_infty$$
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 $mathbbR^n$, 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
$$
left(int_a^b varphi^2n mathrm d alpharight)^1/n leqslant left(int_a^b M^2n mathrm d alpha right)^1/n = M^2 left(int_a^b mathrm d alpha right)^1/n = M^2 (alpha(b) - alpha(a))^1/n to M^2 [n to infty].
$$

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,
$$
forall varepsilon >0, (M - varepsilon)^2 leqslant underline lim left( int_a^b varphi^2n mathrm d alpharight)^1/n leqslant overline lim left( int _a^b varphi^2n mathrm d alpharight)^1/n leqslant M^2.
$$
Therefore the limit exists and equals $M^2$. $blacktriangleright$