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I want to prove if following inequality holds: $$\int_0^1(f')^2\ dx\geq f^2(1)-f^2(0)$$ where $f$ is a function in $H^1([0,1])$ satisfying $\int_0^1f \ dx=0$. It is actually a one dimensional case of following inequality $$\int_{D}|\nabla f|^2\geq \int_{\partial D}f^2$$ where $D$ is the 2-dimensional unit disc and $f$ is a function in sobolev space $H^1(D)$ with mean value zero, i.e., $\int_{D}f=0$. First I'd like to clarify that the above inequality might be not ture, but I have a feeling it's true and I couldn't find a counterexample. Any example and explanation will be helpful.

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It is indeed true.

Here is a proof specific to the 1D case:

$$f(1)-f(0)=\int_0^1 f'(x) dx \leq \int_0^1 |f'(x)| dx$$

and as $f(1)+f(0)=(2\cdot 1 -1) f(1)-(2 \cdot 0-1) f(0)$ with $g(x)=(2x-1) f(x$: $$f(1)+f(0)=\int_0^1 g'(x) dx = \int_0^1 (2x-1) f'(x) dx +\int_0^1 2 f(x) dx = \int_0^1 (2x-1) f'(x) dx $$ and as $|2x-1| \leq 1$ for $x \in[0,1]$: $$f(1)+f(0) \leq \int_0^1 |f'(x)| dx$$

Multiplying both inequalities and using Jensen:

$$f(1)^2-f(0)^2 \leq \left( \int_0^1 |f'(x)| dx \right)^2 \leq \int_0^1 |f'(x)|^2 dx$$


Note that such inequalities are often called trace inequalities instead of Poincaré inequalities (if you need further information).

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