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Se considera functia \( f:[0,1]\to\mathbb{R} \), derivabila, cu derivata continua pe \( [0,1] \). Sa se arate ca daca \( f(1/2)=0 \), atunci

\( \int_0^1(f^{\prime}(x))^{2}dx\geq 12\left(\int_0^1f(x)dx\right)^{2} \).

Cezar Lupu, Tudorel Lupu

Use Fubini theorem
\( \int_{0}^{1}f(x)dx=\int_{0}^{1}(\int_{1/2}^{x}f^{\prime}(t)dt)dx=\int_{0}^{1}k(t)f^{\prime}(t)dt \)

where \( k(t)=-t \) on \( [0;1/2] \) and \( k(t)=1-t \) on \( [1/2;1] \).

With Cauchy-Schwarz
\( (\int_{[0,1]}f)^2 \leq (\int_{[0;1]}k^2)(\int_{[0;1]}f^{\prime2}) \)

and
\( \int_{[0;1]}k^2=\int_{[0;1/2]}t^2dt+\int_{[1/2;1]}(1-t)^2dt=1/24+1/24=1/12 \)

Your inequality with "24" is valid only for small values of 24
More seriously, the second "=" is wrong.
The question if 12 is the best constant remains.

aleph wrote:Your inequality with "24" is valid only for small values of 24
More seriously, the second "=" is wrong.
The question if 12 is the best constant remains.

Yes, the best constant is 12. The problem is not that easy as it seems at first glance.
My solution (which is elementary) is based on Cauchy-Schwarz inequality in the following way:

\( \int_0^{1/2}(f^{\prime}(x))^{2}dx\cdot\int_0^{1/2}x^{2}dx\geq\left(\int_0^{1/2}xf^{\prime}(x)\right)^{2} \)
and
\( \int_{1/2}^{1}(f^{\prime}(x))^{2}dx\cdot\int_{1/2}^{1}(1-x)^{2}dx\geq\left(\int_{1/2}^{1}(1-x)f^{\prime}(x)\right)^{2} \).

By adding these two inequalities and using the elementary inequality \( \alpha^2+\beta^2\geq\frac{(\alpha+\beta)^{2}}{2} \) and integration by parts, we get the desired result. \( \qed \)