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$f:[0,1]\to \mathbb R$ , be continuous function then prove that $$\int_0^1f^2(x)dx\geq \biggl(\int_0^1|f(x)| \biggr) ^2$$

I tried this for $x^2$

For that above is true

But I checked following proof Which is complete opposite to above. Proving the Cauchy-Schwarz integral inequality in a different way

Please help me to find that where is I am making wrong ?

Any help will be appreciated

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  • $\begingroup$ It's not? Just let $g=1$ for your result. $\endgroup$ – user608030 Dec 6 '18 at 15:19
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The inequality that you wrote is not related to the Cauchy-Schwartz inequality. This last inequality states (in this context) that, if $f,g\in C^1\bigl([0,1]\bigr)$, then$$\left(\int_0^1f(x)g(x)\,\mathrm dx\right)^2\leqslant\left(\int_0^1f^2(x)\,\mathrm dx\right)\left(\int_0^1g^2(x)\,\mathrm dx\right).$$If you choose $g=f$, you get the trivial inequality$$\left(\int_0^1f^2(x)\,\mathrm dx\right)^2\leqslant\left(\int_0^1f^2(x)\,\mathrm dx\right)^2$$or$$\int_0^1f^2(x)\,\mathrm dx\leqslant\int_0^1f^2(x)\,\mathrm dx.$$This in no way contradicts what you are supposed to prove.

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  • $\begingroup$ ohh I am applying wrong result. Thanks A lot .You are always helping me. $\endgroup$ – MathLover Dec 6 '18 at 15:37
  • $\begingroup$ ohh Sorry Sir I forget.... $\endgroup$ – MathLover Dec 6 '18 at 15:40
  • $\begingroup$ Sir ,Please can you give some hint so that I can prove above theorem $\endgroup$ – MathLover Dec 6 '18 at 15:43
  • $\begingroup$ Apply the Cauchy-Schwarz inequality to the functions $\lvert f\rvert$ and $1$. $\endgroup$ – José Carlos Santos Dec 6 '18 at 15:45
  • $\begingroup$ Thanks a lots Sir... $\endgroup$ – MathLover Dec 6 '18 at 15:47

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