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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 $$|f(t)| \leq c \left( \int_0^1 |f'(t)|^2 dt \right)^{1/2}$$

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    Forgive me if this is a stupid question, but what does this have to do with $f''$?2012-09-24
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    Also, 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

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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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    Perfect! ...... (Maybe for "Cauchy-Schwarz" for integrals it is fair to give credit also to Bunyakowsky, who apparently was the first to understand a point "obvious" to us nowadays, namely, that the same idea applies to _integrals_, too).2012-09-25
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    @paulgarrett *Justice* is done :)2012-09-25
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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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    In the first sentence, do you mean "total variation of $f\in W^{1,1}$" (i.e., functions with first-order distributional derivative being in $L^1$)?2012-09-25
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    @LVK: indeed. Thanks! It is fixed.2012-09-25