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Any idea why does $$\lim_{n\to\infty}\left(\int_{a}^{b}|f\left(x\right)|^{n}dx\right)^{\frac{1}{n}}=\max_{\left[a,b\right]}f\left(x\right)$$ for $f:[a,b]\to\mathbb{R}$ continuous function?

Simon S
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user114138
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    I suspect you mean the absolute value, $\max |f(x)|$. Can you show that $\lim_{n\to\infty}\sqrt[n]{a^n + b^n} = \max(a,b)$ (for $a, b \geq 0$). Now can you generalize to the Riemann sum? – Simon S Nov 09 '15 at 13:30
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    For the more general case: http://math.stackexchange.com/questions/242779/limit-of-lp-norm – Simon S Nov 09 '15 at 13:38

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Let $x_0 \in [a,b]$ be such that $|f(x_0)| = \max_{x \in [a,b]} |f(x)|$. Now $$ \left( \int_a^b |f(x)|^n \, dx \right)^{1/n} \leq (b-a)^{1/n} |f(x_0)| \to |f(x_0)| $$ as $n \to \infty$.

Also, for all $\varepsilon > 0$ there exists $\delta > 0$ such that $|f(x)| \geq |f(x_0)| - \varepsilon$ for all $x \in (x_0 - \delta, x_0 + \delta)$. Thus \begin{align} \left( \int_a^b |f(x)|^n \, dx \right)^{1/n} & \geq \left( \int_{x_0 - \delta}^{x_0 + \delta} |f(x)|^n \, dx\right)^{1/n} \\ & \geq \left( \int_{x_0 - \delta}^{x_0 + \delta} (|f(x_0)| - \varepsilon)^n \, dx \right)^{1/n} \\ &= (2 \delta)^{1/n} (|f(x_0)| - \varepsilon) \\ & \to |f(x_0)| - \varepsilon \end{align} as $n \to \infty$. Thus the limit is between $|f(x_0)| - \varepsilon$ and $|f(x_0)|$ and the claim follows since $\varepsilon$ was arbitrary.

desos
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