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In the uniform Boundedness theorem it is written $||T_nx|| \le c_x$ and $||T_n|| \le c$

My confusion : what is the difference between $||T_nx|| $ and $||T_n||? $

My thinking : In real analysis we take $f_n(x)=f_n$ so here im confusing that

Are both $||T_nx|| $ and $||T_n|| $ are same?

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    $\begingroup$ $f_n(x) = f_n$ is not true. One is a real number, the other is a function. $\endgroup$ Apr 28, 2021 at 15:18

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They are not the same. $\|T_n\|$ is the operator norm of $T_n$, but $\|T_n x\|$ is the norm of the specific vector $T_n(x)$, where $x$ is an element of the domain of $T_n$.

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$\|T_n\|=\sup_{\|x\|\leq 1} \|T_n(x)\|$

of some $T_n$.

$\|T_n x\|$ is the norm of the image $T_n(x)$.

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In other words, the statement $\| T_n x \| \le c_x$ reads $$ \forall \ x \in E \quad \exists \ c \ge 0 \quad \forall \ n \in \mathbb{N} \quad \| T_n x \| \le c, $$ while the statement $\| T_n \| \le c$ is equivalent to $$ \exists \ c \ge 0 \quad \forall \ x \in E \quad \forall \ n \in \mathbb{N} \quad \| T_n x \| \le c. $$ The latter is obviously stronger. However, Banach-Steinhaus theorem shows they are actually equivalent if $E$ is a Banach space.

And of course, $\| \cdot \|$ here means two different things, as Ben Grossmann pointed out.

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