Consider the space $\mathcal C([0,1])$ of all continuous functions from $[0,1]$ into $\Bbb R$. A norm which is natural to use in this space is the sup norm: $\|f\|_\infty=\sup|f|$; another one is the integral norm: $\|f\|_1=\int_0^1|f|$.
Are $\bigl(\mathcal C([0,1]),\|\cdot\|_\infty\bigr)$ and $\bigl(\mathcal C([0,1]),\|\cdot\|_1\bigr)$ homeomorphic?
My guess is that they are not, but I am unable to prove it. It is clear that these metrics are not equivalent. And, of course, since $\bigl(\mathcal C([0,1]),\|\cdot\|_\infty\bigr)$ is a complete metric space, whereas $\bigl(\mathcal C([0,1]),\|\cdot\|_1\bigr)$ isn't, there is no bijection $f\colon\bigl(\mathcal C([0,1]),\|\cdot\|_\infty\bigr)\longrightarrow\bigl(\mathcal C([0,1]),\|\cdot\|_1\bigr)$ such that both $f$ and its inverse are uniformly continuous, but this doesn't prove the impossibility of the existence of a homeomorphism.
Best Answer
If we embed $(C(X),\|\cdot\|_1)$ into its Banach space completion $E$, then if $C(X)$ were a $G_\delta$ in $E$ then any $f \in E\setminus C(X)$ would contradict Baire's theorem for $E$ as $f + C(X)$ and $C(X)$ are disjoint dense $G_\delta$'s in $E$. So in fact $E=C(X)$. (An old argument by Mazur).
Conclusion: $(C(X),\|\cdot\|_1)$ is not topologically complete (aka Čech-complete). So it cannot be homeomorphic to $(C(X),\|\cdot\|_\infty)$, which is.
The Kadec-Anderson theorem says that if $X$ is a completely metrisable, separable locally convex TVS, then $X \simeq \Bbb R^\omega \simeq \ell_2$, topologically. So the non-completeness is the obstacle.