Maximum value of $|\sum_{n=1}^\infty \frac{g\left(\frac1n\right)}{2^n}|$ Let's consider expression:
$$ L = \left|\sum_{n=1}^\infty \frac{g\left(\frac1n\right)}{2^n}\right|$$
for $$g \in C([0,1]) \quad\text{such that}\quad \int_0^1|g(t)| dt \le 1.$$
I want to find supremum of $L$ under such conditions on $g$. For me the biggest value of $L$ which statisfies my conditions will be taken for $g = 1$. Then $$L = \sum_{n=1}^\infty\frac{1}{2^n} = 1.$$
However, I'm not sure how to prove that there is no other function that will exceed value of $L$.
Could you give me a hand with justifying so?
 A: Note that the p-norm on $C([0,1])$ is defined as $||f||_p:=(\int_0^1|f(t)|^pdt)^{1/p}$ for $p \in [1, \infty)$ and $||f||_\infty=\sup_{t \in [0,1]}|f(t)|$ for $p= \infty$. So basically you know that $||g||_1 \leq 1$.
EDIT: This is wrong in general: (I mixed something up)
You can prove that: $||f||_r \leq ||f||_s $ if $s \leq r$. This implies $||g||_\infty \leq ||g||_1 \leq 1$ . So:
$$L=|\sum_{n=1}^\infty \frac{g(1/n)}{2^n}|\leq \sum_{n=1}^\infty |\frac{g(1/n)}{2^n}| \leq  \sum_{n=1}^\infty\frac{||g||_\infty}{2^n} \leq \sum_{n=1}^\infty\frac{||g||_1}{2^n} \leq \sum_{n=1}^\infty\frac{1}{2^n} =1$$
This doesn't follow:  This means that $L$ can never be higher than 1
EDIT 2: Adding to @Thomas idea here is an example:
Let $M>0$ and consider: $$f(x)=\frac{M}{1+((x-1/2)M \pi)^2}$$
$f\in C([0,1]), f \geq 0, f(1/2)=M$ and
$$\int_0^1|f(t)|dt=\int_0^1f(t)dt=\frac{2}{\pi}\arctan(M\pi/2) \leq 1 $$
If you plot $f$ you can see that the graph gets very large at 1/2 and drops off very fast. This means for $g=f$, $L\geq M/4$, so $L$ can get as big as you want.
A: Given the condition on $g$ it is not difficult to find, for every $M>0$,  a $g>0$ such that:
$g(1/2) > M$
$\int_0^1 g(t)dt=1$
Just think of a delta function centered on 1/2. Told another way, the fact that the integral of $g$ is limited does not say much about its maximum value.
Now it is clear that the addend at $n=1$ is larger than $M/2$. Therefore $L$ can take arbitrarily large values.
