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How to find all $f \in V$ such that $(\int_0^{\ln3}e^x f(x) dx)^2 \leq 2 \int_0^{\\ln 3} e^x [f(x)]^2 dx $

$V$ is a inner product space where $\langle f,g \rangle = \int_0^{\ln3}e^x f(x) g(x)dx$

I know this is a Cauchy-schwarz inequality and I'm able to prove it, but how to find f?

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All $f\in V$ will satisfy this inequality. It is known as Jensen's inequality. Note that $ \frac12\int_0^{\log(3)}e^x\,\mathrm{d}x=1\tag{1} $ Thus, Jensen says that $ \left(\frac12\int_0^{\log(3)}f(x)\,e^x\,\mathrm{d}x\right)^2\le\frac12\int_0^{\log(3)}f(x)^2e^x\,\mathrm{d}x\tag{2} $ which is $\frac14$ of the inequality in question.

Since $e^{x/2}$ is bounded above and below on $[0,\log(3)]$, every function in $L^2[0,\log(3)]$ is also in $V$.

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    @xiamx: If your question was supposed to be "for which $f$ does equality hold?" then that is different. Then you need to look at $ \frac12\int_0^{\log(3)}\left(f(x)-\langle f,f\rangle^{1/2}\right)^2e^x\,\mathrm{d}x =\int_0^{\log(3)}f(x)^2\,e^x\,\mathrm{d}x-\langle f,f\rangle^{1/2}\int_0^{\log(3)}f(x)\,e^x\,\mathrm{d}x $ to see that equality can only happen when $f$ is constant.2012-12-05