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I am stuck on a particular problem. I consider the operator $$A:\mathbf{L}^2[-1,1]\longrightarrow\mathbf{L}^2[-1,1]:f\mapsto Af$$ where I define $Af(t) = t^2f(t)$, and I want to calculate its norm. I already have the following partial results:

  • Since $t^4 \leq 1$,$\forall f\in\mathbf{L}^2[-1,1]$: $$\|Af\|_2^2 = \displaystyle\int\limits_{-1}^1 t^4|f(t)|^2dt \leq \displaystyle\int\limits_{-1}^1 |f(t)|^2dt = \|f\|_2^2$$ so $\|A\|\leq 1$;
  • On the other hand, if I consider $f$ to be the constant $\dfrac{1}{\sqrt{2}}$-function, then obviously $\|f\|_2 = 1$ and it's easy to verify that $$\|Af\|_2^2 = \dfrac{1}{2}\displaystyle\int\limits_{-1}^1 t^4dt = \dfrac{1}{2}\left[\dfrac{t^5}{5}\right]_{-1}^1 = \dfrac{1}{5}$$ So certainly $\|A\|\geq \dfrac{1}{\sqrt{5}}$.

Now what exactly is $\|A\|$?

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    $\begingroup$ Try $f$ equal to the characteristic function of $[1-1/n,1]$ suitability normalised. $\endgroup$
    – Ruy
    Nov 4, 2020 at 18:24
  • $\begingroup$ Yes, that one does the trick. \|A\| = 1 $\endgroup$
    – Werner
    Nov 9, 2020 at 16:33

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