Is This Linear Functional Bounded? In the space of $L^1[0,1]$, is the following linear functional $I：L^1[0,1]\rightarrow \mathbb{R}$:
$$
I(f) = \int_0^1 x^3f(x)dx
$$
bounded on this section of the norm ball $\{f:\|f\|_1 = 1 \text{ and } f(x) > 0, \forall x\}$?

Update: If $I$ is bounded, then the following optimization problem must have a solution:
$$
\max_f I(f)\\
s.t. \|f\|_1 = 1, f > 0,\\
 \int_0^1xf(x)dx = a,\\
 \int_0^1x^2f(x)dx = b.
$$
What is the solution to this problem?
 A: First of all, something stronger:

Lemma Let $X$ and $Y$ normed spaces. Then $T:X\to\mathbb{Y}$ is bounded (in the classical sense: there is $M>0$ such that $\|Tx\|\le
 M\|x\|$ for all $x\in X$) if and only if $T$ maps bounded sets into
  bounded sets.
Proof Suppose $T$ bounded. Let $A\subseteq X$ a bounded set. Then, there is $N>0$ such that $\|x\|\le N$ for all $x\in A$. Thus
  $\|Tx\|\le\|T\|\|x\|\le\|T\|N$, so $T(A)$ is a bounded set on $Y$.
For the reciprocal, let $B=\{x\in X\,|\,\|x\|\le 1\}$. Then $B$ is
  bounded, so $T(B)$ is bounded on $Y$. That is, there is $N>0$ such
  that $\|Tx\|\le N$ for all $x$ with $\|x\|<1$. So $T$ is bounded.

Now, since the set $\{f\,|\,\|f\|=1\mbox{ and }f(x)>0\}$ is bounded, if we proove that $I$ is bounded, done. But
$|I(f)|=\left|\int_0^1x^3f(x)dx\right|\le\int_0^1|x^3||f(x)|dx\le\int_0^1|f(x)|dx=\|f\|$, so $I$ is bounded.
Thus $I$ is bounded on $\{f\,|\,\|f\|=1\mbox{ and }f(x)>0\}$ (on both senses: $f$ is continuous in this set and there is an $N>0$ such that $|I(f)|\le N$ for all $f$ in that set).
