Not all norms are equivalent in an infinite-dimensional space

Simple example of two non-equivalent norms on infinitely-dimensional space:

Consider space of all contnuously differentiable functions $X = C^1 [0,1]$. Then equipping it with the norm: $$ \|f \|_{C^1} = \sup \limits_{x \in [0,1]} |f| + \sup _{x \in [0,1]} |f'| $$ gives us a complete space (Banach space), but if we consider norm: $$ \|f \|_\infty = \sup \limits_{x \in [0,1]} |f| $$ Then the normed space is not complete, hence the norms are not equivalent.

Another way to see it is to consider sequence $f_n(x) = \frac 1n \sin (2 \pi n x)$, which is convergent to $0$ function in $\| \cdot \|_\infty$: $$ \| f_n \|_\infty = \frac 1n \rightarrow 0 $$ norm, but not convergent to $0$ in $ \| \cdot \|_{C^1}$: $$ \| f_n \|_{C^1} = \frac 1n + 1$$

Consider the two spaces $L^{p_1}(-1, 1), L^{p_2}(-1, 1)$ with $1\le p_1<p_2 < \infty$. Let $f \in L^{p_1}\cap L^{p_2}(-1, 1)$. For all $\lambda\ge 1$ define $$ f_\lambda(x)=\lambda^\frac1{p_1}f(\lambda x).$$ Then you have $$\tag{1} \|f_\lambda\|_{p_1}=\|f\|_{p_1} $$ and $$\tag{2} \|f_\lambda\|_{p_2} = \lambda^{\frac{p_2-p_1}{p_1p_2}}\|f\|_{p_2}$$ If the two norms were equivalent on $L^{p_1}\cap L^{p_2}$ you would have, by definition, $$ c\|f_\lambda\|_{p_2} \le \|f_\lambda\|_{p_1}\le C\|f_\lambda\|_{p_2}$$ but (1) and (2) show that this is not possible (to see why, let $\lambda \to \infty$).