Consider the real normed vector space $C([0,1])$ - the space of continuous functions $x: [0,1] \to \mathbb{R}$, with the sup norm (uniform convergence). Consider the linear operator $f: C([0,1]) \to \mathbb{R}$ defined by $f(x) = \lim_{n \to \infty} \int_0^1 x(t^n)dt$. Is this operator continuous? I tried doing the following:
For each $t \in [0, 1)$, $x(t^n)$ converges (pointwise) to $x(0)$, and for each $n \in \mathbb{N}$, $t \in [0, 1]$, $|x(t^n)| \leq ||x||$. So, by the Dominated Convergence Theorem, $f(x) = lim_{n \to \infty} \int_0^1 x(t^n)dt = \int_0^1(\lim_{n \to \infty}x(t^n))dt = x(0)$ - and this implies $|f(x)| = |x(0)| \leq ||x||$. So $f$ is continuous. Is this ok?