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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?

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Try to use Holder inequality, note that you supremum is achievable by functions $f\in L^3[-1,0]$ – Norbert Feb 19 '12 at 22:10
up vote 1 down vote accepted

You can bound $|\phi(f)|$ in terms of $\bigl(\int_0^1|f(x)^3|\,dx\bigr)^{1/3}$ using Hölder's inequality. Then just use $$ \Bigl(\int_0^1|f(x)^3|\,dx\Bigr)^{1/3}\le\Bigl(\int_{-2}^2|f(x)^3|\,dx\Bigr)^{1/3}. $$

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Since it is homework consider that $$\int_{0}^{1}e^x f(x-1)=\int_{-1}^{0}e^{x+1}f(x)dx=\int_{-2}^{2}e^{x+1}I_{[-1,0]}(x)f(x)dx$$

Where $I_{[-1,0]}(x)=1$ if $x\in [-1,0]$ and is zero otherwise.

Use now that $L^{3/2}\equiv(L^{3})^{*}$ and $T:L^{3/2}\rightarrow(L^{3})^{*}$ given by

$T(f)g=\int_{-2}^{2}f(x)g(x)dx$ is an isometry then just evaluate $||e^{x+1}I_{[-1,0]}||_{3/2}$.

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