Wrong defined/boundless $g \rightarrow\sum_{n = 1}^\infty\frac{g(\frac1n)}{2^n}$

Question

Let's consider:

$$f\colon (C[0,1], \Vert\cdot \Vert_1) \ni g\rightarrow \sum_{n=1}^\infty\frac{g(\frac1n)}{2^n}$$

I'm trying to check if this object is well defined.

where $\Vert f \Vert_1 = \int_0^1|f(t)|dt$

Let's take function $g(x) = e^{\frac1x}$

Then this sum above will be turned into:

$$f(g) = \sum_{n=1}^\infty (\frac{e}{2})^n$$

But this sequence cannot converge, becuase $(\frac e2)^n \rightarrow \infty, n \rightarrow \infty$.

So - I just found a function that $f(\text{that_function}) = \infty$.

Am I proved that funcional $f$ is wrongly defined or I just proved that $f$ it's boundless ?

My queries are derived from the fact that in my way of thinking I haven't used $\Vert \cdot \Vert_1$ anywhere.

Answer 1

If $g$ belongs to $V=(\mathcal C([0,1], \mathbb R), \Vert \cdot \Vert_1)$, $g$ is continuous and therefore bounded on the compact segment $[0,1]$. The series

$$f(g) = \sum_{i=1}^\infty\frac{g(\frac1n)}{2^n}$$ is absolutely convergent and convergent. Conclusion: $f$ is defined on $V$.

Note: the norm $\Vert \cdot \Vert_1$ has nothing to do regarding the question of $f$ being well defined or not. The norm would obviously play a role on questions like the continuity of the functional $f$.