# Norm of a certain operator in a Hilbert space

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 $$\|A\|$$?

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