# Show that $(C[0,1],\|\cdot\|_{L^1})$ is not a Banach space [duplicate]

Let the space $(C[0,1],\|\cdot\|_{L^1})$ where $$\forall f \in C([0,1]); \|f\|_{L^1}= \int^1_0 |f(x)| \, dx$$ show that $(C[0,1],\|\cdot\|_{L^1})$ is not a Banach Space

Def of Banach space is a complete norm space.

def of complete if it is a cauchy seq in norm space then it converges in the norm space.

Guessing that a seq that might work would be something like a trig function that would be cauchy but it will not converge in the space (either by not converging or not being continous) or somhow fail to be a norm space. Not sure what route to take at this point. Will try to play with obvious cauchy seq and see what happens.

A hint would be appreciated thanks.

$$f_n(x) = \begin{cases} 0 & \text{if } 0 \le x \le \dfrac 1 2 - \dfrac 1 {2^n}, \\[6pt] 1 & \text{if } 1 \ge x \ge \dfrac 1 2 + \dfrac 1 {2^n} \\[6pt] \cdots & \cdots \end{cases}$$ and in the interval with endpoints $\dfrac 1 2 \pm \dfrac 1 {2^n}$ the graph is a straight line and $f$ is continuous. Then $f_n\in L^1$ and $f_n$ is continuous, and $f_n$ converges in $L^1$ (and also pointwise) to a function that is in $L^1$ but is not continuous, but does not converge in $L^1$ to any function that is continuous.