Let $f$ is absolutely continuous function on $[0,1]$, $f(0)=0$ and $f' \in L^2[0,1]$. Would you help me to prove that there is constant $c$ such that $|f(t)| \leq c \left( \int_0^1 |f'(t)|^2 dt \right)^{1/2}$
$f'$ is in $L^2[0,1]$
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1Also, as Byron Schmuland pointed out in a comment to an answer I just deleted, if $c$ is meant to be independent of $f$ then you should say so! – 2012-09-24
2 Answers
Since $ f(t) = f(0) + \int_0^t f'(x) dx = \int_0^t f'(x) dx $ we have $ \lvert f(t) \rvert \leq \int_0^t \lvert f'(x) \rvert dx \leq \int_0^1 1 \cdot\lvert f'(x) \rvert dx $ Now, using Cauchy-Bunyakovsky-Schwarz inequality in $L^2[0, 1]$, we conclude $ \int_0^1 1 \cdot\lvert f'(x) \rvert dx\leq \left(\int_0^1 1^2\cdot dx \right)^{1/2} \left(\int_0^1 \lvert f'(x) \rvert^2 dx \right)^{1/2} = \left(\int_0^1 \lvert f'(x) \rvert^2 dx \right)^{1/2} $
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2@paulgarrett *Justice* is done :) – 2012-09-25
The total variation of $f\in W^{1,1}[0,1]$ is given by $ \operatorname*{Var}_{[0,1]}\,f=\int_0^1\left|f^\prime(t)\right|\,\mathrm{d}t\tag{1} $ $(1)$ implies that for any $x,y\in[0,1]$, $ \left|f(x)-f(y)\right|\le\int_0^1\left|f^\prime(t)\right|\,\mathrm{d}t\tag{2} $ Furthermore, for any convex $\phi$, Jensen's Inequality says $ \phi\left(\int_0^1g(t)\,\mathrm{d}t\right)\le\int_0^1\phi(g(t))\,\mathrm{d}t\tag{3} $ Using the special cases of $y=0$, $g=\left|f'\right|$, and $\phi(x)=x^2$ yields $ \left|f(x)\right|^2\le\int_0^1\left|f^\prime(t)\right|^2\,\mathrm{d}t\tag{4} $
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0@LVK: indeed. Thanks! It is fixed. – 2012-09-25