For $f \in L^1[0,1]$ let $x_n = \int_0^1 f(t) t^n \: dt$, and let $T(f) = \{ x_n \}$. I want to show that the operator $T$ is a bounded linear operator mapping $L^1[0, 1]$ to $c_0$ (the space of real convergent sequences that converge to $0$) and determine the norm of $T$.
I have tried to go about this in a direct manner but haven't had any luck. For a bounded linear operator I am using the standard definition:
Let $X$ and $Y$ be normed linear spaces. A linear operator $T: X \to Y$ is bounded if there exists an $M \geq 0$ such that $||T u|| \leq M ||u||$ for all $u \in X$.
Lastly it is worth noting that I am considering the standard norm of $c_0$. That is to say $$||(x_1, x_2, x_3, \cdots)|| = \sup \{ |x_n| : n \in \mathbb{N} \}.$$
Best Answer
Let me complement QuantumSpace's answer by showing that $\|T\| = 1$ but $T$ does not attain its norm.
We shall first show that $\|T\| = 1$. The inequality $\|T\| \leq 1$ has already been shown, so I will show the other inequality. For this define, for $n \in \mathbb{N}$, $f_n(t) = n \chi_{[1-\frac{1}{n},1]}$. That is $f_n(t) = n$ for $t \in [1-\frac{1}{n},1]$ and $f_n(t) = 0$ otherwise. It follows that $$\|f_n\|_1 = \int_0^1 |f_n(t)| dt = \int_{1-\frac{1}{n}}^1 n dt = n \frac{1}{n} = 1.$$ Further, $$\|Tf_n\| \geq \int_0^1 |f_n(t)t| dt = n \int_{1-\frac{1}{n}}^1 t dt \geq n \int_{1-\frac{1}{n}}^1 (1-\frac{1}{n}) dt = n \frac{1}{n}(1-\frac{1}{n}) = 1- \frac{1}{n}.$$ It follows that $\|T\| \geq 1-\frac{1}{n}$. As $n \in \mathbb{N}$ was arbitrary, $\|T\| \geq 1$ and we are done.
Now we will show that $T$ does not attain its norm. We have the following inequalities for any $f \in L_1$: $$\|Tf\| \le \sup_n \int_0^1 |f(t)t^n|dt \le \int_0^1 |f(t) t|dt \overset{(*)}{\le} \int_0^1 |f(t)|dt = \|f\|_1.$$
Note that $(*)$ becomes an equality if and only if $f = 0$ a.e. Otherwise we would have $|f|>0$ on a set $M$ of positive measure and thus $|f(t)t| < |f(t)|$ on $M$ which would give us the sharp inequality in $(*)$. It follows that $T$ does not attain its norm.