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Let $\{f_k\}_{k=1}^{\infty}$ be a sequence in $L^p(\Omega,\Sigma,\mu)$ for $1\leq p<\infty$. Suppose $0<c=\inf_k \lVert f_k\rVert_p\leq \sup_k \lVert f_k\rVert_p=C<\infty$ and $f_if_j=0$ for $i\neq j$. Let $(a_k)_{k=1}^{\infty}$ be a sequence of real numbers such that $\sum_k \lvert a_k\rvert^p<\infty$. Then by the completeness of $L^p$, $f=\sum_{k=1}^\infty a_k f_k$ is a function in $L^p(\Omega,\Sigma,\mu)$. Moreover,

$$ \lVert f\rVert _p\leq C(\sum_{k=1}^\infty \lvert a_k\rvert ^p)^{1/p}. $$

And when $p=2$, I can obtain

$$ \lVert f\rVert_2\geq c(\sum_{k=1}^\infty \lvert a_k\rvert^2)^{1/2}. $$

In general, how to prove $$ \lVert f\rVert _p\geq c(\sum_{k=1}^\infty \lvert a_k\rvert^p)^{1/p} $$ ??? I have no way to prove it.

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Let $S_n:=\{x,f_n(x)\neq 0\}$. Then $\mu(S_i\cap S_j)=0$ if $i\neq j$ (this is the key point). We thus have $$\lVert f\rVert^p_p=\sum_{i\geqslant 1}\int_{A_i}|f(x)|^p\mathrm d\mu=\sum_{i\geqslant 1}\int_{A_i}|a_i|^p|f_i(x)|^p\mathrm d\mu=\sum_{i\geqslant 1}|a_i|^p\int_{A_i}|f_i(x)|^p\mathrm d\mu\\=\sum_{i\geqslant 1}|a_i|^p\int_{\Omega}|f_i(x)|^p\mathrm d\mu$$ and the result follows.

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