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For $f,g \in C[a,b]$ let $d_p(f,g)=\left( \int_a^b\left|f(t)-g(t)\right|^pdt\right)^{1/p}$ and $d_{\infty}=\max_{a\leq t \leq b}\left|f(t)-g(t)\right|$. Prove that $\lim _{p\rightarrow \infty}d_p(f,g)=d_{\infty}(f,g)$.

I'm trying to prove that $d_{\infty}(f,g)\leq d_p(f,g) \leq (b-a)^{1/p}d_{\infty(f,g)}$. The second part of the inequality I already proved it, but the first one is giving me a hard time. Could you please help me?

Thanks!

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  • $\begingroup$ Let $a = 0$ and $b = 1$, and let us pose that $f(x)=x^{\frac{1}{p}}$ and $g = 0$, such that $d_{\infty}(f,g) = 1$ and $d_p(f,g)^p = \int_{0}^1 tdt = \frac{1}{2}$, and $1 > \frac{1}{2^p}$, thus the left inequality is false. $\endgroup$ – Hermès Oct 3 '16 at 16:59
  • $\begingroup$ Ok, then how can I prove the limit? $\endgroup$ – Gabrielita Oct 3 '16 at 17:03

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