I do not want to argue against

**Opalg** because I am a noob compared to him in analysis

, but I offer a suggestion.

I think this problem is better solved using the concepts of limsup's and liminf's.

Let $\displaystyle \ell$ be the length of the interval. Let $\displaystyle I_n = \left( \smallint_a^b f^n \right)^{1/n}$.

Opalg hath shown us that $\displaystyle \delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq I_n \leq m \cdot \ell^{1/n}$.

Then he says $\displaystyle m-\epsilon \leq \lim I_n \leq m \implies I_n = m$ by $\displaystyle \epsilon \to 0^+$.

But we need to know the limit exists.

So a way around this problem is to perhaps say first take $\displaystyle \epsilon \to 0^+$ and then $\displaystyle n\to \infty$.

But the problem is $\displaystyle \delta$ depends on $\displaystyle \epsilon$ and so how we know that $\displaystyle \delta^{1/n} \not \to 0$ as $\displaystyle n\to \infty$?

Maybe I am missing something but I think that if we use limsup's and liminf's the solution is much smoother. Because $\displaystyle \delta^{1/n}\cdot \ell^{1/n}\cdot (m-\epsilon) \leq I_n$ and so $\displaystyle m-\epsilon \leq \liminf I_n \implies m \leq \liminf I_n$. However, $\displaystyle I_n \leq m \implies \limsup I_n \leq m$. Thus, we have shown $\displaystyle \liminf I_n \geq \limsup I_n$ but $\displaystyle \liminf I_n \leq \limsup I_n$. This means that $\displaystyle I_n$ is a convergent sequence with $\displaystyle \lim I_n = \liminf I_n = \limsup I_n = m$.

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