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Suppose I have proved for all $f$ positive with $\lVert f \rVert_{L^2} = 1$ that $$\int f^2\log(f) \leq -C_1\log(\epsilon) + \epsilon\lVert \nabla f \rVert_{L^2}^2.$$ I want to prove that for all $f$ with norm not necessarily $1$ that $$\int f^2\log\left(\frac{|f|^2}{\lVert f \rVert_{L^2}^2}\right) \leq -C_1\log(\epsilon)\lVert f \rVert_{L^2}^2 + \epsilon\lVert \nabla f \rVert_{L^2}^2$$

How do I do this? I Tried taking $f=\frac{f}{\lVert f \rVert_{L^2}}$ but I just can't get the log in left hand side right.

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    $\begingroup$ Use $\log(x^2)=2\log(x)$ $\endgroup$ Jul 13 '14 at 20:43
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We can let $f \mapsto \frac{f}{\lVert f \rVert_{L^2}}$ and as @mathematician suggested, we can multiply by $2 \times \lVert f \rVert_{L^2}$.

Note that we get a factor of two on the right hand side so we do not get the desired inequality but a close enough one.

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