Show that $T: L^2([0,1]) \to L^2([0,1])$, $f(x) \mapsto x \cdot f(x)$ is bounded.

Question

Let $T: L^2([0,1]) \to L^2([0,1])$, $f(x) \mapsto x \cdot f(x)$. I want to show that $T$ is bounded by finding an upper bound on its norm $\| T \|$.

What I've tried

Using Cauchy-Schwarz we have \begin{align} \| T \|^2 & = \sup_{\| f \|_2 = 1} \| T f \|^2 = \sup_{\int_{0}^{1} |f(x)|^2 dx = 1} \int_{0}^{1} x^2 | f(x) |^2 dx \\ & \le \sup_{\int_{0}^{1} |f(x)|^2 dx = 1} \sqrt{\int_{0}^{1} x^4 dx} \sqrt{\int_{0}^{1} | f(x) |^4 dx} \\ & = \frac{1}{\sqrt{5}}\sup_{\int_{0}^{1} |f(x)|^2 dx = 1} \sqrt{\int_{0}^{1} | f(x) |^4 dx} \end{align} I am not sure if the next step is correct: $$ \sup_{\int_{0}^{1} |f(x)|^2 dx = 1} \sqrt{\int_{0}^{1} | f(x) |^4 dx} \le \sup_{\int_{0}^{1} |f(x)|^2 dx = 1} \sqrt{\left(\int_{0}^{1} | f(x) |^2 dx\right)^{2}} = 1. $$ If the above step is incorrect, how can we continue or is there a neater way? Hints are greatly appreciated.

Answer 1

The solution is easier than you think: $$ ||Tf||^2 =\int_0^1 |xf(x)|^2 dx =\int_0^1|x|^2|f(x)|^2 dx \le \int_0^1 |f(x)|^2 dx =||f||^2. $$ So $||T||\le 1$, and by considering functions of the form $n\mathcal{X}_{[1-1/n, 1]}$ you can see that in fact $||T||=1$.

Answer 2

Actually, $\sup\left\{\int_0^1 \lvert f(x)\rvert^4\,dx\,:\,f\in L^2[0,1]\land\int_0^1 \lvert f(x)\rvert^2\,dx=1\right\} =\infty$. However, you can bound $\int_0^1x^2\lvert f(x)\rvert^2\,dx$ with another instance of Holder's inequality (case $p=1,q=\infty$ instead of $p=q=2$):

$$\int_0^1x^2\lvert f(x)\rvert^2\,dx\le\lVert x^2\rVert_\infty\lVert \lvert f\rvert^2\rVert_1=\lVert f\rVert^2$$

where $\lVert \bullet\rVert_\infty$ is the norm of $L^\infty[0,1]$.