$(C_{[0;1]}, \Vert . \Vert_1)$ is not a Banach space

Question

I'm going to prove that $C_{[0;1]}$ is not a Banach space w.r.s to the norm $\Vert x \Vert_1 = \int_{0}^{1} |x(t)| dt$ by consider the series $\sum_{n=1}^{\infty}x_n$ where $x_{n}(t)= t^{n} \cdot \sqrt{1-t}$.

By MCT we shall easy check that this series is absolutely convergence. Indeed $$ \begin{array}{ll}\sum_{1}^{\infty}\Vert x_n \Vert_1 & = \sum_{1}^{\infty}\int_{0}^{1}x_{n}(t)dt = \int_{0}^{1} \sum_{1}^{\infty} x_n(t)dt=\int_{0}^{1}\sum_{1}^{\infty}t^n\sqrt{1-t}\, dt \\ &= \int_0^1 \frac{t}{\sqrt{1-t}}dt = \frac{4}{3} \end{array} $$ But I'm stuck in proving this series is not convergence in $(C_{[0;1]}, \Vert . \Vert_1)$. I think that if this series is convergence in $(C_{[0;1]}, \Vert . \Vert_1)$ to $x$, it may convergence to $x$ pointwise, but I can't prove this.

Thank you for your help.

Answer 1

Notice that, if $g(t)=\sum_1^\infty x_n(t)$, then, for $t\in (0,1)$ $$ g(t)= \frac{\sqrt{1-t}}{1-t}. $$ You can prove this as you did, using the fact that we have a geometric series. On the other hand, since $x_n(0)=0$ for all $n$, we have $g(0)=0$, which is impossible since the sum representation above forces $g(0)=1$. Therefore the series doesn't converge in your space.

Answer 2

Your series are convergence in this norm indeed, and you have already proven it. For an example to show the noncompleteness, consider fn(x) to be a function such that it equal to 1 when x < 1/2, it equal to 0 when x > 1/2 + 1/n, and it is linear when 1/2 < x < 1/2 + 1/n. If such fn converges to a function f in the norm above, then view it as L1 we get a subsquence converges a.e. to f, thus f must be the function equal to 1 when x < 1/2 and equal to 0 when x > 1/2 a.e., and cannot be continuous.