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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Consider functional $$ F:X\to\mathbb{R}:x\mapsto\int\limits_{0}^{1}x(t)dt $$ and prove that $M=\mathrm{Ker}(f)$ and $\Vert F\Vert=1$.
Recall that $$ \mathrm{dist}(x_0,M)=\frac{|F(x_0)|}{\Vert F\Vert} $$ Here you can find the proof of this fact.
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 $$ \forall x\in X\quad\Vert x\Vert_\infty=1\implies|F(x_0)|<1 $$
Conclude that $d(x_0,M)<1$
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could you please write it more detailed – 89085731 Oct 07 '12 at 03:39
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@89085731 Which part of my answer require more detail? – Norbert Oct 07 '12 at 07:31
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||F||=1 I don't know – 89085731 Oct 07 '12 at 13:01
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As for this part use estimation $$\left|\int\limits_0^1 x(t)dt\right|\leq\int\limits_{0}^{1}|x(t)|dt\leq\int\limits_{0}^{1}\Vert x\Vert_\infty dt=\Vert x\Vert_\infty\int\limits_{0}^{1}dt=\Vert x\Vert_\infty$$ – Norbert Oct 07 '12 at 13:24
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@89085731 Is there anything unclear now? – Norbert Oct 07 '12 at 15:44
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@Norbert But this only tells that $|F|\le 1$. For showing it equals to 1, we have to show that if $|F|=\alpha<1$ then there exist an element $y\in X, |y|=1$ such that $|\int_0^1y(t)dt|>\alpha$. – Sachchidanand Prasad Aug 28 '17 at 07:18
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@SachchidanandPrasad, indeed consider functions $y_n(t)=\max(1,nt)$ and choose $n$ so that $F(y)>\alpha$. – Norbert Aug 28 '17 at 07:53