A semi-norm on $\ell_p$ spaces and its relation to the usual norm Let $1\leq p <\infty,$ $A$ be an infinite set, and  $\theta=\{\theta_a: \ell_p(A) \to \Bbb R| a\in A\}$ be a family of bounded linear functional which satisfies $\sum_{a\in A}|\theta_a(f)|^p<\infty$ for all $f\in \ell_p(A).$
Define $\|f\|_{\theta}:= \big(\sum_{a\in A}|\theta_a(f)|^p\big)^{1\over p}.$
My question is about the relation between the two values $\|f\|_{\theta}$ and $\|f\|.$
It is clear that $\|\cdot\|_{\theta}$ is a semi-norm on $\ell_p(A).$ Also, $\|f\| \leq M \|f\|_{\theta}$  not necessarily holds; (for example, we can consider all $\theta_i$ s equal to zero). Now, is there some $M\geq 0$ such that  $\|f\|_{\theta} \leq M \|f\|$ holds for all $f\in\ell_p(A)?$
 A: Define a linear operator $T:\ell^p(A)\to\ell^p(A)$ by $f\mapsto T(f)$, where
$$T(f)(a):=\theta_a(f)\text{ for all }a\in A.$$
If we show that $T$ is bounded, then we are done, since then we would have $\|T(f)\|\leq\|T\|\cdot\|f\|$ for all $f\in\ell^p(A)$ and it is clear that
$$\|T(f)\|=\bigg(\sum_{a\in A}|T(f)(a)|^p\bigg)^{1/p}=\bigg(\sum_{a\in A}|\theta_a(f)|^p\bigg)^{1/p}=\|f\|_\theta.$$
Note that the operator does indeed take values in $\ell^p(A)$, since by assumption we have $\|f\|_\theta<\infty$ for all $f\in\ell^p(A)$, i.e. $\|T(f)\|_{\ell^p(A)}<\infty$, i.e. $T(f)\in\ell^p(A)$.
To show that $T$ is bounded, we use the closed graph theorem: Let $(f_n)\subset\ell^p(A)$ be a sequence with $\|f_n\|_{\ell^p(A)}\to0$ and assume that $T(f_n)\to g$ in $\ell^p(A)$. If we show $g=0$, we are done. Fix $a\in A$. We have
$$|g(a)|\leq|g(a)-T(f_n)(a)|+|T(f_n)(a)|=\big(|g(a)-T(f_n)(a)|^p\big)^{1/p}+|\theta_a(f_n)|\leq$$
$$\leq\bigg(\sum_{b\in A}|g(b)-T(f_n)(b)|^p\bigg)^{1/p}+|\theta_a(f_n)|=\|g-T(f_n)\|_{\ell^p(A)}+|\theta_a(f_n)|\to0,$$
since by assumption $T(f_n)\to g$ in $\ell^p(A)$ and $f_n\to0$ and $\theta_a$ is continuous, hence $\theta_a(f_n)\to0$. As $a\in A$ was arbitrary, this shows $g=0$. By the closed graph theorem, we have that $T$ is bounded.
Comment: There were comments that this is immediate by the uniform boundedness principle. Even though the closed graph theorem is very closely related to the UBP, I don't see what the right application of the UBP is to make this an immediate consequence. I do agree though, that the above is pretty much just the standard application of the closed graph theorem.
