Let $T_n:X:=C([0,1])\rightarrow C([0,1])=:Y$ be a sequence of linear operators. Suppose:

  • $T_n$ is bounded
  • $(Tf)(x):=\lim\limits_{n\to\infty}(T_nf)(x)$ exist for every $f\in C([0,1])$ and $x\in [0,1]$
  • $Tf\in C([0,1])$ for every $f\in C([0,1])$

Show that $T$ is linear and bounded. It is true that the last assumption follows from the first and second ?

So, if $Af\in C([0,1])$ then the boundedness and linearity follows form uniform boundedness principle, since it gives us that $\mathbb{T}:=\{T_n\}$ is uniformly continuous and hence if $Tf$ exists in $C([0,1])$ then take $U=\overline{U}\in \mathcal{O}_0^Y$. We can find $V\in\mathcal{O}_0^X$ s.th. for any $n\ge 1 \ T_n(V)\subset U $. Hence for any $f\in V $:

$T(f)=\lim_\limits{n\to \infty}T_nf\in Y$. So, $T(V)\subset U$.

But I don't no idea how to do if we omit last assumption.


No, the last assumption does not follows from the first two one. To see this consider operators $T_n f = f\left(x^n \right).$


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