Proving that limit of a sequence $\{f_k\}\in\mathcal C([0,1],\Bbb R$ is not in $\mathcal C([0,1],\Bbb R)$

Question

So, I was given the task to prove that the sequence $\{f_k\}\in\mathcal C([0,1],\Bbb R)$ defined by: $$f_k(x)= \begin{cases} 1, & \text{if $x \in [0, \frac 12[$} \\ -2^k(x-\frac 12)+1 , & \text{if $x \in [\frac 12, \frac 12 + \frac {1}{2^k}[$} \\ 0, & \text{if $x \in [\frac 12 + \frac {1}{2^k},1]$} \\ \end{cases} $$ does not converge under the $||•||_1$ norm (I proved already that it is Cauchy).

I'm almost done, the sequence would be of functions whose values are 1 on $[0,\frac 12[$, then a line until it intersects the x-axis on $\frac 12 + \frac {1}{2^k}$, then 0 for the rest of the interval. As $k$ grows, the magnitude of the slope of the line increases, it intersects the x axis before the previous function, and it's value is 0 on a longer interval. So, i saw that this sequence tends to a function that is 1 on $[0, \frac 12]$, and 0 in $[ \frac 12, 1]$, and that cannot be since this function is not continous, thus the sequence does not converge.

So, my proof goes like this:

Assume $f_k \rightarrow f \in \mathcal C([0,1],\Bbb R)$, now assume that $\exists x_0 \in [ \frac 12, 1] \text { such that } f(x_0) \neq 0$, then, since $f$ is continous, $\exists \delta, \forall x \in [x_0- \delta, x_0+\delta], f(x) \neq 0$ $$\int_{x_0- \delta}^{x_0+ \delta} |f(x)|dx \gt 0$$ Now I have to prove somehow that this implies the existence of a lower bound for $||f_k-f||_1$, Impliying that f can not be the limit. Then, we have that the previous assumption is wrong, so $\forall x \in [ \frac 12, 1] f(x_0)=0$, this contradicts the countinuity of $f$, proving by contradiction that such an f does not exist.

Answer 1

I don't think you should focus on the norm part, but rather on that, if it converged to some $f \in \mathcal{C}([0,1],\mathbb{R})$, $f$ would be continuous.

Hint: Suppose such a limit $f$ exists. What is the value of $f$ for $x\in[0,1/2)$? And for $x\in (1/2,1]$? Can $f$ be assigned a value at $x=1/2$ that makes it continuous?

Answer 2

Another way (although here we work inside the complete space $B([0,1],\mathbb R)$:

Clearly $f_n(x)$ converges to $0$ if $x<1$ and to $1$ if $x=1$. To $f_n$ converges pointwise to the function $g:[0,1]\rightarrow \mathbb R$ defined by $g(x)=0$ if $x\neq 1$ and $g(1)=1$. If $f_n$ converged uniformly to a function then it would also converge pointwise to the same function, since $g\not\in \mathcal C([0,1],\mathbb R)$ we conclude that $f_n$ does not converge.

Note: in fact $f_n$ does not converge uniformly to $g$, one way to see this rather easily is by the Arzela-Ascoli theorem.