Consider ${ \mathcal{C}([0,1]) },$ the space of continuous functions on ${ [0,1] }.$ For ${ p \geq 1 }$ it has ${ p- }$norm ${ \lVert f \rVert _{p} := (\int \vert f \vert ^p ) ^{\frac{1}{p}} }.$ It also has sup norm ${ \lVert f \rVert _{\infty} := \sup _{t \in [0,1]} \vert f(t) \vert }.$
For now we’ll only consider ${ \lVert .. \rVert _{1}, }$ ${ \lVert .. \rVert _{2} }$ and ${ \lVert .. \rVert _{\infty} }.$
${ \lVert .. \rVert _2 }$ is a bit special because it is the norm induced by inner product ${ \langle f,g \rangle := \int _{0} ^{1} f(t) g(t) dt }.$
Here ${ \lVert f \rVert _1 }$ ${ \leq \lVert f \rVert _2 }$ ${ \leq \lVert f \rVert _{\infty} },$ as ${ \lVert f \rVert _1 }$ ${ = \int _{0} ^{1} \vert f \vert \cdot 1 }$ ${ \leq \sqrt{ \int _{0} ^{1} \vert f \vert ^2} \sqrt{ \int _{0} ^{1} 1 ^2 } }$ ${ = \lVert f \rVert _2 }$ ${ \leq \sqrt{ \lVert f \rVert _{\infty} ^2 \cdot (1-0) } }$ ${ = \lVert f \rVert _{\infty} }.$ But the norms are pairwise inequivalent :
Cauchy sequences in these normed spaces
Turns out ${ \mathcal{C}([0,1]) }$ is complete wrt ${ \lVert .. \rVert _{\infty} }$ but not complete wrt ${ \lVert .. \rVert _1, \lVert .. \rVert _2 }.$
Completeness wrt sup norm :
Let ${ (f _n ) }$ be a Cauchy seq in ${ (\mathcal{C}[0,1], \lVert .. \rVert _{\infty} ). }$
For each ${ t \in [0,1] },$ seq ${ (f _n (t) ) }$ is Cauchy (from ${ \vert f _m (t) - f _n (t) \vert }$ ${ \leq \lVert f _m - f _n \rVert _{\infty} }$) and hence converges to a point, which we’ll call ${ f(t) }.$
Firstly ${ f \in \mathcal{C}[0,1] }.$
Let ${ p \in [0,1] }$ and ${ \epsilon \gt 0 }.$
There is an ${ N }$ such that ${ \lVert f _m - f _n \rVert _{\infty} \lt \epsilon }$ for all ${ m \geq n \geq N }.$ So for every pair ${ m \geq n \geq N }$ and ${ t \in [0,1] },$ we have ${ \vert f _m (t) - f _n (t) \vert \lt \epsilon }.$ Fixing ${ n \geq N, t \in [0,1] }$ and letting ${ m \to \infty },$ we see ${ {\color{green}{\vert f(t) - f _n (t) \vert \leq \epsilon} } }$ for every choice of ${ n \geq N }$ and ${ t \in [0,1] }.$
Pick ${ \delta \gt 0 }$ such that ${ t \in [0,1], \vert t - p \vert \lt \delta }$ implies ${ \vert f _{N} (t) - f _{N} (p) \vert \lt \epsilon }.$
Now ${ \vert f(t) - f(p) \vert }$ ${ \leq {\color{green}{\vert f(t) - f _{N} (t) \vert} } }$ ${ + \vert f _N (t) - f _N (p) \vert }$ ${ + {\color{green}{ \vert f _N (p) - f (p) \vert } } }$ ${ \lt 3 \epsilon }$ whenever ${ t \in [0,1] }$ and ${ \vert t - p \vert \lt \delta },$ as needed.
Also from the green proposition in above proof, ${ \lVert f _n - f \rVert _{\infty} \to 0 ,}$ as required.
Incompleteness wrt ${ \lVert .. \rVert _1 }$ :
Consider seq ${ (g _n) }$ as in the above pic. Here ${ \int _{0} ^{1} \vert g _n (t) - \frac{1}{\sqrt{t}} \vert dt }$ ${ = \int _{0} ^{\frac{1}{n}} \frac{1}{\sqrt{t}} dt - \sqrt{n} \frac{1}{n} }$ ${ = \frac{1}{\sqrt{n}} },$ ie ${ \int \vert g _n - \frac{1}{\sqrt{t}} \vert = \frac{1}{\sqrt{n}} }.$
So ${ \lVert g _m - g _n \rVert _1 }$ ${ = \int \vert g _m - g _n \vert }$ ${ \leq \int \vert g _m - \frac{1}{\sqrt{t}} \vert }$ ${ + \int \vert g _n - \frac{1}{\sqrt{t}} \vert }$ ${ = \frac{1}{\sqrt{m}} + \frac{1}{\sqrt{n}}, }$ making ${ (g _n ) }$ Cauchy wrt ${ \lVert .. \rVert _1 }.$ But it has no limit in ${ (\mathcal{C}[0,1], \lVert .. \rVert _1) }.$
Suppose there were a ${ g \in \mathcal{C}[0,1] }$ with ${ \lVert g _n - g \rVert _1 }$ ${ = \int _0 ^1 \vert g _n (t) - g (t) \vert dt \to 0 }.$ We already have ${ \int _0 ^1 \vert g _n (t) - \frac{1}{\sqrt{t}}\vert dt }$ ${ = \frac{1}{\sqrt{n}} \to 0 }.$
So ${ \int _0 ^1 \vert g(t) - \frac{1}{\sqrt{t}} \vert dt }$ ${ \leq \int _0 ^1 \vert g(t) - g _n (t) \vert dt }$ ${ + \int _0 ^1 \vert g _n (t) - \frac{1}{\sqrt{t}} \vert dt \to 0 },$ giving ${\underline{ \int _0 ^1 \vert g(t) - \frac{1}{\sqrt{t}} \vert dt = 0. } }$
Now ${ \vert g(t) - \frac{1}{\sqrt{t}} \vert }$ is ${ 0 }$ on ${ (0,1] },$ because if it were ${ \neq 0 }$ at some ${ c \in (0,1] }$ then by continuity there is a ${ \delta \gt 0 }$ such that it is ${ \neq 0 }$ on ${ (c-\delta, c] },$ giving ${ \int _{0} ^{1} \vert g(t) - \frac{1}{\sqrt{t}} \vert dt \gt 0 }.$
So ${ g(t) = \frac{1}{\sqrt{t}} }$ on ${ (0,1] }.$ Especially ${ g }$ is unbounded, a contradiction.
Incompleteness wrt ${ \lVert .. \rVert _2 }$ :
Consider a seq ${ (h _n) }$ in ${ \mathcal{C}[0,1] },$ where ${ h _n (t) }$ is ${ t ^{-\frac{1}{4}} }$ on ${ [\frac{1}{n}, \infty ) }$ and ${ n ^{\frac{1}{4}} }$ on ${ [0, \frac{1}{n} ] }.$
So ${ \left( \int _0 ^1 \vert h _n (t) - t ^{-\frac{1}{4}} \vert ^2 dt \right) ^{\frac{1}{2}} }$ ${ = \left( \int _0 ^{\frac{1}{n}} (t ^{-\frac{1}{4}} - n ^{\frac{1}{4}} ) ^2 dt \right) ^{\frac{1}{2}} }$ ${ \leq \left( \int _0 ^{\frac{1}{n}} (t ^{-\frac{1}{2}} + n ^{\frac{1}{2}}) dt \right) ^{\frac{1}{2}} }$ ${ = \sqrt{3} n ^{-\frac{1}{4}}, }$ that is ${ \left( \int \vert h _n - t ^{-\frac{1}{4}} \vert ^2 \right) ^{\frac{1}{2}} \leq \sqrt{3} n ^{-\frac{1}{4}} . }$ So ${ \lVert h _m - h _n \rVert _2 }$ ${ \leq \left( \int \vert h _m - t ^{-\frac{1}{4}} \vert ^2 \right) ^{\frac{1}{2}} }$ ${ + \left( \int \vert h _n - t ^{-\frac{1}{4}} \vert ^2 \right) ^{\frac{1}{2}} }$ ${ \leq \sqrt{3} m ^{-\frac{1}{4}} + \sqrt{3} n ^{-\frac{1}{4}} ,}$ making ${ (h _n) }$ Cauchy wrt ${ \lVert .. \rVert _2 }.$
Proceeding as in the ${ \lVert .. \rVert _1 }$ case, we’re done.