# Showing that the operator $T$ is a bounded linear operator mapping $L^1[0, 1]$ to $c_0$

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

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.

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 $$\|T\| \le 1$$.

• 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? Commented Mar 6, 2022 at 15:46
• @AnIsomorphicTeen I added some details. Commented Mar 6, 2022 at 15:52
• Thank you for your help! Commented Mar 6, 2022 at 16:11
• 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. Commented Mar 6, 2022 at 17:44
• @KeeperOfSecrets Yes you are right. Commented Mar 6, 2022 at 18:41