# Norm of this operator

I'm trying to figure out the norm $||\phi||$ of the functional $\phi: L^3[-2,2] \to \mathbb{C}$ defined by

$$\phi(f)=\int_{0}^{1}e^xf(x-1)dx$$

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)|| &= |\int_{0}^{1}e^x f(x-1)dx|\\ &=|\int_{-1}^0e^{x+1}f(x)dx \\ &=|\int_{-2}^{2}e^{x+1}I_{[-1,0]}f(x)dx| \\ &\leq\int_{-2}^{2}|e^{x+1}I_{[-1,0]}f(x)|dx \\ &\leq(\int_{-2}^{2}|e^{x+1}I_{[-1,0]}|^{\frac{3}{2}}dx )^{\frac{2}{3}}(\int_{-2}^{2}|f(x)|^3dx )^{\frac{1}{3}} \\ &=(\int_{-1}^{0}|e^{x+1}|^{\frac{3}{2}}dx )^{\frac{2}{3}}||f||_3\\ &=(\int_{0}^{1}|e^{x}|^{\frac{3}{2}}dx )^{\frac{2}{3}}||f||_3\\ &=(\frac{2}{3}(e^{\frac{3}{2}}-1))^{\frac{2}{3}}||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 $f\in L^3[-1,0]$ –  userNaN Feb 19 '12 at 22:10

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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