Norm of Position Operator in $L^2[0,1]$

Question

I was wondering what is the norm of the position operator $Xf(x)=xf(x)$ in $L^2[0,1]$. I have two different results.

1. The first one is the simplest and the reasonable: $$||X|| \overset{||f(x)||=1}{=} \sup||xf(x)||=\sup||x||=1, $$ since $x\in[0,1]$.

2. The second method is the usual one I have always applied for L^2 opeartors:

$$||Xf(x)||^2=\left(\int_0^1xf(x)\text{d}x\right)^2\le\left|\int_0^1x^2\text{d}x\right| \left|\int_0^1f^2(x)\text{d}x\right|=\frac13||f(x)||^2 \qquad \implies \qquad ||X||=\frac{1}{\sqrt{3}}.$$

Both are different, and I am not able to find the mistake in the second method. Can you help me?

Thank you in advance :)

Answer 1

The "mistake" is in both methods.

In the first one, you are mixing norms. The result is correct, though.

In the second, you have $\|Xf\|_2^2=\int_0^1 x^2 f^2(x)$, and you cannot apply Cauchy-Schwarz.

The usual calculation is $$ \|Xf\|=\left(\int_0^1 x^2 f^2(x)\,dx\right)^{1/2}\leq\left(\int_0^1 f^2(x)\,dx\right)^{1/2}=\|f\|, $$ so $\|X\|\leq 1$. Further for each $\delta>0$ you can find $f$ with $\|Xf\|\geq(1-\delta)\|f\|$, proving that $\|X\|=1$.