# Alternative definition of the $L^p$ norm

Is it true for $1 \leq p<\infty$ that for $f \in L^p(\mathbb R)$ and $q$ is such that $1/p+1/q=1$, $$\|f\|_p= \sup \left\{\int_\mathbb R fg : g\in L^q(\mathbb R), \|g\|_q \leq 1 \right\} ?$$

The integral is with respect to the Lebesgue measure.

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By Hölder's inequality, $LHS \geqslant RHS$, and for the choice $g:=f^{\frac qp}$, assuming without loss of generality that $f\geqslant 0$, we can see that $LHS\leqslant RHS$.

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Thank you for the answer. If I take the sup over functions with some smoothness, say $g \in C^k$ do I still get the same $p$-norm or something strictly smaller? –  Cantor Mar 28 '13 at 14:49
Smooth functions with compact support are dense in $L^q$ for $1\leqslant q<\infty$, so for $1<p<\infty$, it still work. –  Davide Giraudo Mar 28 '13 at 14:51
Do I need absolute values inside the integral (i.e. $\sup\int|fg|$)? Also, could you write out the details of $LHS \leq RHS$? It seems to me that $g$ should be taken to be $\frac{f^{p/q}}{\|f\|_p^{p/q}}$ –  Cantor Mar 28 '13 at 19:37

Let $f = \chi_{[0,2]}$. Then $||f||_2 = \sqrt 2$. On the other hand, if $g = 100 \chi_{[0,2]}$, then $\int fg = 200$, a contradiction.

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I think I need to take the sup over $\|g\|_q \leq 1$. –  Cantor Mar 28 '13 at 14:31
That sounds right. –  user66345 Mar 28 '13 at 14:31
@Cantor, now Davide's answer is right. –  user66345 Mar 28 '13 at 14:41
As it stands it's not correct or at least missing some crucial details. Do you care to fill in the details? –  Cantor Mar 29 '13 at 11:19