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It is a problem in my homework. Let $$ X = \{x \in C[0,1] : x(0) = 0\} $$ with norm $\Vert\cdot\Vert_\infty$. Denote $$ M =\left\{ x \in X : \int\limits_0^1 x(t)=0\right\} $$ If $\Vert x_0\Vert_\infty=1$ and $x_0\in X$ how to prove that $d(x_0,M)<1$

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What have you tried? –  F M Oct 7 '12 at 0:13

1 Answer 1

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  1. Consider functional $$ F:X\to\mathbb{R}:x\mapsto\int\limits_{0}^{1}x(t)dt $$ and prove that $M=\mathrm{Ker}(f)$ and $\Vert F\Vert=1$.

  2. Recall that $$ \mathrm{dist}(x_0,M)=\frac{|F(x_0)|}{\Vert F\Vert} $$ Here you can find the proof of this fact.

  3. For $x\in C([0,1])$ with $\sup_{t\in[0,1]}|x(t)|1$ the integral $\left|\int_0^1 x(t)dt\right|$ will attain its maximum for the functions $x(t)=1$ and $x(t)=-1$. But you don't have this functions in the space $X$. So show that $$ \forall x\in X\quad\Vert x\Vert_\infty=1\implies|F(x_0)|<1 $$

  4. Conclude that $d(x_0,M)<1$

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could you please write it more detailed –  Joe Oct 7 '12 at 3:39
    
@89085731 Which part of my answer require more detail? –  userNaN Oct 7 '12 at 7:31
    
||F||=1 I don't know –  Joe Oct 7 '12 at 13:01
    
As for this part use estimation $$\left|\int\limits_0^1 x(t)dt\right|\leq\int\limits_{0}^{1}|x(t)|dt\leq\int\limits_{0}^{1}\Vert x\Vert_\infty dt=\Vert x\Vert_\infty\int\limits_{0}^{1}dt=\Vert x\Vert_\infty$$ –  userNaN Oct 7 '12 at 13:24
    
@89085731 Is there anything unclear now? –  userNaN Oct 7 '12 at 15:44

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