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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 $$||(x_1, x_2, x_3, \cdots)|| = \sup \{ |x_n| : n \in \mathbb{N} \}.$$

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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.

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Given $f\in L^1[0,1]$, let $$x_n:= \int_0^1 f(t)t^n dt.$$

We want to show that $\{x_n\}_{n=1}^\infty$ converges to $0$.

However, this is easy. Note that $f(t)t^n \to 0$ for almost every $t \in [0,1]$ (namely, for all $t \in [0,1)$). Further, note that $|f(t)t^n| \le |f(t)|$ for all $t \in [0,1]$, so we are in a situation where we can apply the dominated convergence theorem to deduce that $$\lim_{n \to \infty} x_n = \lim_{n \to \infty} \int_0^1 f(t)t^n dt = \int_0^1 0 dt = 0$$ as desired. Hence, $T$ maps $L^1[0,1]$ into $c_0$, as desired.

Note now that $$\sup_n |x_n| \le \sup_n \int_0^1 |f(t)t^n|dt \le \|f\|_1$$ which shows that $\|T\| \le 1$.

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  • $\begingroup$ Thank you for your input! I have two questions still though. 1) How can I deduced that T is a bounded linear operator? 2) What would the norm of T be? $\endgroup$ Commented Mar 6, 2022 at 15:46
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    $\begingroup$ @AnIsomorphicTeen I added some details. $\endgroup$
    – J. De Ro
    Commented Mar 6, 2022 at 15:52
  • $\begingroup$ Thank you for your help! $\endgroup$ Commented Mar 6, 2022 at 16:11
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    $\begingroup$ I have some doubts about the computation of the norm. Your $f$ is not in $L_1$. And even if it was, you would still have to show that $\|f\|_{L_1} \leq 1$ to get $\|T\| \geq 1$. Actually, I believe that this $T$ does not attain its norm. $\endgroup$ Commented Mar 6, 2022 at 17:44
  • $\begingroup$ @KeeperOfSecrets Yes you are right. $\endgroup$
    – J. De Ro
    Commented Mar 6, 2022 at 18:41

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