I'm supposed to show that the following operator

$$T:L^{1}(0,2)\rightarrow L^{1}(0,2),\ (Tf)(x):=\int_{0}^{x}tf(t)\ \text{d}t$$

is well defined. The first thing I'm supposed to do is to make sure that $(Tf)(x)$ makes sense, i.e.,

$$\int_{0}^{x}|tf(t)|\ \text{d}t<\infty\ \ \text{for all }f\in L^{1}(0,2),\ x\in [0,2]$$

Here's my idea: I can manipulate a little with the above expression

$$\int_{0}^{x}|tf(t)|\ \text{d}t\leq \int_{0}^{2}|t||f(t)|\ \text{d}t=\int_{0}^{2}t|f(t)|\ \text{d}t$$

Because we're integrating from 0 to $x$ and over $|tf(t)|$, then the first inequality in the above expression makes sense. Furthermore, if we integrate from 0 to $2$, then $|t|=t$, since we have positive limits of integration.

But after here, I get stuck. Any hints on how to proceed would be highly appreciated!

  • 1
    $\begingroup$ As for the convergence of the integral, notice that $2|f(t)|$ is integrable and dominates $tf(t)$ on $[0, 2]$. To show that $Tf$ is $L^1$, Fubini's theorem may help you. $\endgroup$ – Sangchul Lee Apr 14 '18 at 23:12
  • 2
    $\begingroup$ I think we can safely say that $\int_0^2 t|f(t)|dt \leq \int_0^2 2|f(t)|dt .$ $\endgroup$ – Dalamar Apr 14 '18 at 23:14
  • $\begingroup$ That makes sense. Thanks for the help! $\endgroup$ – James Apr 15 '18 at 16:13

Hint: $x\in[0,2]$ and so $t\in[0,x]\subset[0,2$. Thus $$\int_{0}^{x}|tf(t)|\ \text{d}t\leq \int_{0}^{2}|t||f(t)|\ \text{d}t\le2\int_{0}^{2}|f(t)|\ \text{d}t<\infty.$$


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