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The problem is to show that the sequence of functions $f_n(t) = $ \begin{cases} 0 & 0 \leq t \leq \frac{1}{2} -\frac{1}{n}\\ \frac{1}{2}+\frac{n(t-\frac{1}{2})}{2}& \frac{1}{2} -\frac{1}{n} < t < \frac{1}{2} +\frac{1}{n} \\ 1 & \frac{1}{2} +\frac{1}{n} \leq t \leq 1 \end{cases}

is a Cauchy sequence in $L^2[0, 1]$ but not in C[0,1] (i.e., using the uniform norm).

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I have proved that in for continuous function [0,1] it is not Cauchy:

Let $m \leq n$, consider $|| f_m(x) - f_n(x)||$, as $n \rightarrow \infty$, $||f_n(x)|| \rightarrow \frac{1}{2}$ where $x = \frac{1}{2}$. And since the function becomes an uncontinuous fucntion, it is not Cauchy.

But for the $L^2[0,1]$ space, I know $||f_m(x) - f_n(x)||_2 = =\int_{0}^1 (f_m(x) - f_n(x))^2dx$. I don't know how should I proceed.

Can anyone help me to check if for $C[0,1]$ is a right aprroach and how to check for $L^2[0,1]$?

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let be $n\geq 1$, we have:

$$ \lVert f_{2n} - f_{n}\rVert_{\infty} \geq \left| f_{2n}\left(\frac{1}{2} + \frac{1}{2n}\right) -f_n\left(\frac{1}{2} + \frac{1}{2n}\right)\right| = \frac{1}{4} $$ so $f_n$ is not Cauchy in $C^{\infty}[0,1]$

Let be $h\in L^2[0,1]$: $$h(x) = \begin{cases} 0 & \text{if } 0 \leq x < \frac{1}{2}\\ 1 & \text{otherwise} \end{cases} $$

we have $$ \lVert f_m - f_n\rVert \leq \lVert f_m - h\rVert + \lVert f_n - h\rVert$$

so it suffice to prove $\lVert f_n - h\rVert_{L^2} \to 0$

P.S.: we have to prove $f_n$ is Cauchy in $L^2$, but we know that $L^2$ is complete, so it is natural to search for an $h$ s.t. $\lVert f_n - h \rVert_{L^2} \to 0$

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