The function space $C([a,b])$ is the space of all continuous, real valued functions defined on $[a,b]$ and has the supremum norm ($\|\cdot\|_{\infty}$) associated with it. I came across a book which says that the supremum norm is complete on this space, but the two-norm ($\|\cdot\|_{2}$) isn't.
How does the norm associated with the space determine the completeness? Does the convergence of a Cauchy sequence in the space depend on what kind of normed space it is?
First, I prove the claim that $C([0,1])$ is not complete in the $\|\cdot\|_{2}$ norm defined by $\|f\|_{2} = \left(\int_{0}^{1} |f(x)|^{2} \, dx\right)^{\frac{1}{2}}$.
Consider the function $f : [0,1] \to \mathbb{R}$ given by $$f(x) = \left\{\begin{array}{r l} 0, & x < \frac{1}{2} \\ 1, & x \geq 1 \end{array} \right.$$ $f$ has a jump discontinuity at $\frac{1}{2}$ so it doesn't live in $C([0,1])$. Showing that there's a sequence $(f_{n})_{n \in \mathbb{N}}$ in $C([0,1])$ that converges to $f$ in the $\|\cdot\|_{2}$-norm is tricky since $f$ itself isn't in $C([0,1])$. We could appeal to the natural embedding of $C([0,1])$ in $L^{2}([0,1])$, but I think it's more instructive to work with $C([0,1])$ on its own terms.
For each $n \in \mathbb{N} \setminus \{1,2\}$, define $f_{n} : [0,1] \to \mathbb{R}$ given by $$f_{n}(x) = \left\{ \begin{array}{r l} 0, & x < \frac{1}{2} - n^{-1} \\ \frac{n}{2}\left(x - \frac{1}{2} + n^{-1}\right), & \frac{1}{2} - n^{-1} \leq x \leq \frac{1}{2} + n^{-1} \\ 1, & x > \frac{1}{2} + n^{-1} \end{array} \right.$$ The above is just a verbose way of writing $f_{n}$ is the function that's $0$ until $\frac{1}{2} - n^{-1}$, it goes up linearly to $1$ at $\frac{1}{2} + n^{-1}$, and then it's $1$ thereafter. It's one of many natural ways of approximating the $f$ given above by continuous functions. Notice that $f_{n}(x) \to f(x)$ as $n \to \infty$ whenever $x \in [0,1]$.
I claim $(f_{n})_{n \in \mathbb{N} \setminus \{1,2\}}$ is Cauchy in the $L^{2}$-norm. The easiest way to see this is the following: suppose $\epsilon > 0$ and pick $M \in \mathbb{N} \setminus \{1,2\}$ such that $M^{-1} < \frac{\epsilon}{4}$. If $m,k \geq M$, then $f_{m}$ and $f_{k}$ are equal everywhere except in the interval $[\frac{1}{2} - M^{-1}, \frac{1}{2} + M^{-1}]$, where there difference is bounded, at any rate, by $2$. Thus, $$\left(\int_{0}^{1} |f_{m}(x) - f_{k}(x)|^{2} \, dx \right)^{\frac{1}{2}} = \left(\int_{\frac{1}{2} - M^{1}}^{\frac{1}{2} + M^{-1}} |f_{m}(x) - f_{k}(x)|^{2} \, dx \right)^{\frac{1}{2}} \leq 2 \cdot 2M^{-1} = 4 M^{-1}.$$ This shows $\|f_{m} - f_{k}\|_{2} < \epsilon$ if $m,k \geq M$. By arbitrariness of $\epsilon > 0$, the sequence is Cauchy.
Nonetheless, $(f_{n})_{n \in \mathbb{N} \setminus \{1,2\}}$ does not converge in the $\|\cdot\|_{2}$ norm to any continuous function. I could try to convince you that $f_{n} \to f$ in the $\|\cdot\|_{2}$-norm, but that doesn't quite make sense at this point since I would have to appeal to a larger space where $f$ lives. Instead, I'll prove for you that if $g \in C([0,1])$, then there is a $\delta > 0$ and an $M \in \mathbb{N} \setminus \{1,2\}$ such that $\|f_{n} - g\|_{2} \geq \delta$ if $n \geq M$. The key intuition here: imagine thinking about what the difference $\|f - g\|_{2}$ should be. If $g\left(\frac{1}{2}\right)$ is close to $1$ and $g$ doesn't change very much in the interval $(\frac{1}{2} - \delta, \frac{1}{2} + \delta)$, then we pick up a penalty on the order of $\sqrt{\delta}$ on the left side of $\frac{1}{2}$. The same thing happens if $g\left(\frac{1}{2}\right)$ is close to $0$ and we instead look at the right side of $\frac{1}{2}$.
More formally, suppose first that $g\left(\frac{1}{2}\right) \geq \frac{1}{2}$ (i.e. it's "close" to $1$). Pick a $\delta > 0$ so that $g(x) \geq \frac{1}{3}$ if $x \in (\frac{1}{2} - \delta, \frac{1}{2} + \delta)$. Pick $M \in \mathbb{N} \setminus \{1,2\}$ such that $M^{-1} < \frac{\delta}{2}$. If $n \geq M$, then $f_{n}$ vanishes in the interval $(\frac{1}{2} - \delta, \frac{1}{2} - \frac{\delta}{2})$ and, thus, \begin{align*} \|f_{n} - g\|_{2} &\geq \left(\int_{\frac{1}{2} - \delta}^{\frac{1}{2} - \frac{\delta}{2}} |f_{n}(x) - g(x)|^{2} \, dx \right)^{\frac{1}{2}} \\ &= \left(\int_{\frac{1}{2} - \delta}^{\frac{1}{2} - \frac{\delta}{2}} |g(x)|^{2} \, dx \right)^{\frac{1}{2}} \\ &\geq \frac{1}{3} \sqrt{\frac{\delta}{2}}. \end{align*} This proves the claim when $g\left(\frac{1}{2}\right) \geq \frac{1}{2}$. In the case when $g\left(\frac{1}{2}\right) < \frac{1}{2}$ (intuitively, $g \left(\frac{1}{2}\right)$ is "closer to zero"), repeat the above argument by looking at the right side of $\frac{1}{2}$.
The above shows that completeness of a normed space $(B,\|\cdot\|)$ depends as much on the norm $\|\cdot\|$ as it does on the linear space $B$.
