I'm trying to figure out the norm $\|\phi\|$ of the functional $\phi: L^3[-2,2] \to \mathbb{C}$ defined by
$ \phi(f)=\int_{0}^{1}e^xf(x-1)\mathsf dx$
but am struggling. I can't figure out how to write $\|\phi(f)\|$ in terms of $\|f\|$ because the domain of integration within the definition of this function is $[0,1]$ (or $[-1,0]$ if you change the variable), rather than $[-2,2]$.
Any tips?
EDIT: In light of the answers, I've worked out the following:
$\begin{align} \|\phi(f)\| &= \left|\int_{0}^{1}e^x f(x-1)\mathsf dx\right|\\ &=\left|\int_{-1}^0e^{x+1}f(x)\mathsf dx\right| \\ &=\left|\int_{-2}^{2}e^{x+1}I_{[-1,0]}f(x)\mathsf dx\right| \\ &\leq\int_{-2}^{2}|e^{x+1}I_{[-1,0]}f(x)|\mathsf dx \\ &\leq\left(\int_{-2}^{2}|e^{x+1}I_{[-1,0]}|^{\frac{3}{2}}\mathsf dx\right)^{\frac{2}{3}}\left(\int_{-2}^{2}|f(x)|^3\mathsf dx\right)^{\frac{1}{3}} \\ &=\left(\int_{-1}^{0}|e^{x+1}|^{\frac{3}{2}}\mathsf dx \right)^{\frac{2}{3}}\|f\|_3\\ &=\left(\int_{0}^{1}|e^{x}|^{\frac{3}{2}}\mathsf dx \right)^{\frac{2}{3}}\|f\|_3\\ &=\left(\frac{2}{3}\left(e^{\frac{3}{2}}-1)\right)^{\frac{2}{3}}\right)\|f\|_3 \end{align} $
Am I allowed to pull the integral sign inside as I did in the first inequality when $f$ may be complex valued? And if all this is right, how would I got about showing that this is indeed the minimal constant?