Answer: $L^p$ Norm $\to$ $L^\infty$

Key Idea

Use the Squeeze Theorem: find upper and lower bounds for $\|f\|_p$ that both converge to $\|f\|_\infty$ as $p \to \infty$.

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Proof

Let $M = \|f\|_\infty$.

Upper Bound

Since $|f(x)| \leq M$ a.e., we have:

$\|f\|_p = \left(\int_X |f|^p \, d\mu\right)^{1/p} \leq \left(\int_X M^p \, d\mu\right)^{1/p} = M \cdot \mu(X)^{1/p}$

Since $\mu(X) < \infty$, as $p \to \infty$:

$\mu(X)^{1/p} = e^{\frac{1}{p}\ln \mu(X)} \to e^0 = 1$

Therefore $\displaystyle\limsup_{p \to \infty} \|f\|_p \leq M$.

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Lower Bound

For any $\varepsilon > 0$, define:

$E_\varepsilon = \{x \in X : |f(x)| > M - \varepsilon\}$

By definition of essential supremum, $\mu(E_\varepsilon) > 0$. Then:

$\|f\|_p \geq \left(\int_{E_\varepsilon} |f|^p \, d\mu\right)^{1/p} \geq (M - \varepsilon) \cdot \mu(E_\varepsilon)^{1/p}$

As $p \to \infty$, again $\mu(E_\varepsilon)^{1/p} \to 1$, so:

$\liminf_{p \to \infty} \|f\|_p \geq M - \varepsilon$

Since $\varepsilon > 0$ was arbitrary:

$\liminf_{p \to \infty} \|f\|_p \geq M$

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Conclusion

Combining both bounds:

$M \leq \liminf_{p \to \infty} \|f\|_p \leq \limsup_{p \to \infty} \|f\|_p \leq M$

Therefore:

$\lim_{p \to \infty} \|f\|_p = M = \|f\|_\infty \qquad \blacksquare$

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Why $\mu(X) < \infty$ is needed

If $\mu(X) = \infty$, the upper bound step fails — $\mu(X)^{1/p} \to \infty$ and the argument breaks. (Counterexample: $f \equiv 1$ on $\mathbb{R}$ with Lebesgue measure gives $\|f\|_p = \infty$ for all $p < \infty$ but $\|f\|_\infty = 1$.)