I have a problem solving following: Its a Counterexample to Banach-Steinhaus. Its bounded for every point but not uniformly

We have $X=C^0 (I), ||.||_{L^2}$ with $I = [0,1]$ $||f||_{L^2}= (\int_I |f(t)|^2 dt) ^{1/2}$

Now there are linear Operators $T_n = n \int_0^{1/n} f(t) dt$

with $T_n : X \rightarrow \mathbb{R}$

So we have to show that $sup||Tx||<\infty$ is true but $sup||T||<\infty$ isnt.

I dont even know how to begin here

Sorry i missed the "n" in $T_n$.

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    $\begingroup$ Where is your question? $\endgroup$ – Demophilus Nov 16 '17 at 14:46
  • $\begingroup$ Begin by stating the problem correctly... If $T_n$ is as above then $||T_n||\to0$. $\endgroup$ – David C. Ullrich Nov 16 '17 at 14:47
  • $\begingroup$ The question is, why is it bounded but not uniformly bounded? $T_n$ is exactly what i posted. $\endgroup$ – Joseph S. Nov 16 '17 at 17:15
  • $\begingroup$ The definition of $T_n$ does not make sense. If $T_nf = \int_0^{1/n}f(t)dt$, then $|T_nf| \le \|f\|_{L^2}$, hence the $T_n$ are uniformly bounded... $\endgroup$ – daw Nov 16 '17 at 21:09
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    $\begingroup$ Possible duplicate of Banach-Steinhaus and sequence of functions $\endgroup$ – daw Nov 17 '17 at 12:49

If $f_n$ is $n^{1/2}$ on (0,1/n), 0 on (2/n,] and linear inbetween then $f_n$'s are bounded in $L^2$ norm and $T_n f_n$ is unbounded. These functiosn do not vanish at 0 but it is easy to modify this example so that the new functions vanish at 0.

  • $\begingroup$ I guess i have to show that its in general bounded for all $f_n$ and then pick an example its uniformly unbounded. I know that f is bounded (compact and continous on [0,1]). issue is i dont see the counter example. $sup||T_n||=( \int_I (n \int_0^{1/n} n^{1/2})^2 dt)^{1/2} $ $\endgroup$ – Joseph S. Nov 17 '17 at 7:50

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