# Banach-Steinhaus and sequence of functions

I'm looking at the following problem:

Let $X:=(C^0([0,1]), || \cdot||_{L^2})$, $Y:=\mathbb R$ and $$||f||_{L^2}:=(\int^1_0 |f(t)|^2dt)^{1/2}$$ for $f \in C^0([0,1])$. For $n \in \mathbb N$ we define $T_n:X \rightarrow Y$ as $$T_n(f):=n \int^{1/n}_0 f(t)dt$$ $i)$ The $T_n$ are pointwise bounded linear functionals.

$ii)$ The $T_n$ are not uniformly bounded.

$iii)$ What do $i)$ and $ii)$ imply regarding Banach-Steinhaus theorem and why?

For $iii)$ I got: $i)$ and $ii)$ imply that Banach-Steinhaus does not apply here. The reason is that $X$ is not complete (I showed that using a counterexample). My problem is regarding the other two parts: The $T_n$ are obviously linear functionals, but what about the "bounded" attributes? I have to show something similar for another sequence of functions... If I knew how to show the attribute"pointwise/uniformly bounded" above, then I could solve my other problem as well.

• @Demophilus I don't see how one could use Cauchy-Schwarz. Could you elaborate? – Yasuduck Nov 17 '17 at 0:54
• Sorry, that hint was completely wrong. I posted a solution. – Demophilus Nov 17 '17 at 1:45

We know any $f \in C^0[0,1]$ is bounded, meaning we have $\| f\|_\infty = \sup_{f \in [0,1]} \lvert f(x) \rvert < +\infty$. So for any $f \in C^0[0,1]$ we have $$\lvert T_n(f) \rvert \leq n\int_0^{1/n}\lvert f(t)\rvert dt \leq n\int_0^{1/n}\|f\|_\infty dt = \|f\|_\infty.$$ This means that $T_n$ are pointwise bounded linear functionals.

To show that $T_n$ aren't uniformly bounded, it's sufficient to find a sequence of functions $f_n$ in $C^0[0,1]$ with $\|f_n\|_2=1$ such that $\lvert T_n(f_n)\rvert$ isn't bounded. First we define $$g_n(t) = \left \{ \begin{matrix} 1 & \text{ if } t \leq 1/n \\ 2-nt & 1/n<\text{ if } t \leq 2/n \\ 0 & \text{ else}\end{matrix} \right.$$ Now note that $\|g_n\|_2 =\sqrt{ \frac{4}{3n}}$ so define $f_n = \frac{g_n}{\|g_n\|_2}$. Then we find $$\lvert T_n(f_n) \rvert = n \sqrt{\frac{3n}{4}} \int_0^{1/n} g_n(t) dt =\sqrt{\frac{3n}{4}}.$$ So the functionals $T_n$ are uniformly bounded.