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Let $X\in{l^p}$, with $1\le{p}\le{q}\le{\infty}$ and $X=(x_j)_{j=1}^\infty$ s.t. $\sum_{j=1}^{\infty}{|x_j|^p}<\infty$ (by definition). We choose $X$ s.t $\|X\|_{l^p}=\big(\sum_{j=1}^{\infty}{|x_j|^{p}}\big)^{1/p}=1.$ I am trying to show that $l^p\subset{l^q}$ and $\|X\|_{l^q}\le{\|X\|_{l^p}}.$

My lecturer's proof starts with the fact that:

"Suppose that $X\in{l^p}$ with $\|X\|_{l^p}=1$, and so in particular $|x_j|\le{1}$ for every $j$."

But I can't see how this is true, is there a nice way to show this or is it obvious?

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    $\begingroup$ $1=\sum_n|x_n|^p\geq |x_j|^p$. The inequality is due to removing some non-negative terms. $\endgroup$ – logarithm May 9 at 15:42
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As pointed out before, $$1=\sum_{n=1}^{+\infty}\left\lvert x_n\right\rvert^p=\left\lvert x_j\right \rvert^p+\sum_{n=1,n\neq j}^{+\infty}\left\lvert x_n\right\rvert^p\geqslant \left\lvert x_j\right \rvert^p.$$

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