I'm trying to calculate operator norm in $C[-1,1]$ and $L_1[-1,1]$ for the following:
$$(Af)(x)= \int_{-1}^{1}(3t^2+x)f (t)dt$$
I found that in $C[-1,1]$:
\begin{align} \lVert A \rVert &\leq \max_{x\in [-1,1]} \int_{-1}^{1}\vert 3t^2+x \vert \mathrm {dt} \\ &= \max_{x\in [-1,1]} \left (\int_{-1}^{-\sqrt{-\frac{x}{3}}} (3t^2+x) \mathrm {dt} + \int_{-\sqrt{-\frac{x}{3}}}^{\sqrt{-\frac{x}{3}}} (-3t^2-x) \mathrm {dt} + \int_{\sqrt{-\frac{x}{3}}}^{1} (3t^2+x) \mathrm {dt}\right) \\ &= \max_{x\in [-1,1]} \left(-4\sqrt{\frac{x}{3}} -4x \cdot \sqrt{-\frac{x}{3}} +2+2x \right) \\ &=2 \end{align}
and in $L_1 [-1,1]$:
$$\lVert A \rVert \leq \int_{-1}^{1} \max_{t\in [-1,1]}\vert 3t^2+x \vert \mathrm {dx} = \int_{-1}^{1} (3+x) \mathrm {dx}=6$$
Is this is correct? I would appreciate any help! Thx.