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Let $M=\Vert f{\Vert }_{\infty }$, the essential supremum of $|f|$. If $M=0$, $f=0$ almost everywhere and every Lp norm is zero. Suppose $M>0$. Since $|f|\le M$ almost everywhere and $m((0,1))=1$,For any $a$ with $0<a<M$, the measurable set $A_a=\{x\in (0,1):|f(x)|>a\}$ has positive Lebesgue measure; otherwise $a$ would be an essential upper bound. HenceFor fixed $a$, $m(A_a{)}^{1/p}\to 1$. Thus ${\mathrm{liminf}}_{p\to \infty }\Vert f{\Vert }_p\ge a$. Letting $a\uparrow M$ and combining with the upper bound provesThe positive-measure superlevel sets explain why the Lp norms converge to the supremum norm even when the essential maximum is never attained.

Put $p_j=2{\kappa }^j$, where $\kappa =n/(n-2)>1$. Since $p_j^2/(p_j-1)\le 2p_j$, part (ii), followed by a $p_j$-th root, gives the Moser iteration stepAfter $N$ steps,The geometric series and its differentiated form giveIn particular the exact convergent product isThe printed hint's equality to a single power of $2\kappa$ is not correct in general; its claimed finiteness is correct and the displayed expression supplies the correction.

On a finite-measure domain, Lp norms converge to the supremum norm for a bounded continuous function. Indeed, the upper bound is $\Vert u{\Vert }_p\le |\Omega {|}^{1/p}\Vert u{\Vert }_{\infty }$; for every $a<\Vert u{\Vert }_{\infty }$, the set $\{|u|>a\}$ has positive measure, giving $\Vert u{\Vert }_p\ge a|\{|u|>a\}{|}^{1/p}$. Let $N\to \infty$ in the iteration and use part (i), which gives $\lambda >0$. The Dirichlet eigenfunction supremum estimate isThe eigenvalue exponent is exactly $n/4$. The constant depends only on $n$: the supplied zero-boundary Sobolev inequality has a dimension-only constant, and every product factor above depends only on $\kappa$.