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$f\in C[0,2\pi]$ and $f_\delta := f$ in $[\delta , 2\pi - \delta],f_{\delta}(0)=f_{\delta}(2\pi)=\frac{1}{2}(f(0)+f(2\pi))$ and $f_\delta$ is linear in $[0,\delta]$ and also in $[2\pi - \delta , 2\pi]$

For every $\epsilon > 0$ there exists a $\delta > 0 $ such that : $||f-f_\delta||<\epsilon$.

How to prove this?

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Which norm are talking about? –  Norbert Feb 2 '13 at 12:13
    
the norm induced by : $||f||:= (f,f)^{1/2} = (\int_{0}^{2\pi}f^2)^{1/2})$ –  bakabakabaka Feb 2 '13 at 12:15
    
I think this question raised when you were thinking about Fourier series, but as the question stated it is not related to Fourier series at all –  Norbert Feb 2 '13 at 12:17
    
I tried to correct the title, is it better suited now? –  bakabakabaka Feb 2 '13 at 12:22
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1 Answer

up vote 1 down vote accepted

Since $f=f_\delta$ on $[\delta,2\pi-\delta]$, then $$ \Vert f-f_\delta\Vert= \left(\int\limits_{[0,\delta]}|f(x)-f_\delta(x)|^2 dx+\int\limits_{[2\pi-\delta,2\pi]}|f(x)-f_\delta(x)|^2 dx\right)^{1/2} $$ Let $M=\max\limits_{x\in[0,2\pi]}|f(x)|$, then for all $x\in[0,2\pi]$ we have $|f_\delta(x)|\leq M$ and we conculde $$ \Vert f-f_\delta\Vert\leq\left(\int\limits_{[0,\delta]}(2M)^2 dx+\int\limits_{[2\pi-\delta,2\pi]}(2M)^2 dx\right)^{1/2}=2M(2\delta)^{1/2} $$ So for a given $\varepsilon$ it is enough to set $\delta=\varepsilon^2/(8M^2)$.

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Thank you very much! –  bakabakabaka Feb 2 '13 at 14:30
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