Let $1\leq p<q\leq\infty$
Consider the spaces $X=(C(a,b), \|\cdot\|_{L^q(a,b)})$ and $Y=(C(a,b), \|\cdot\|_{L^p(a,b)})$. I want to show $X$ is continuously embedded in $Y$, but $Y$ is not continuously embedded in $X$.
Proof attempt:
Note an operator between normed spaces is continuous if and only if bounded, so it suffices to show that $\exists C>0$ s.t. $\|x\|_Y\leq C\|x\|_X$
We want the inclusion map $\iota(f) \mapsto f$, to be bounded for $f\in C(a,b)$ where $\|f\|_{L^q}$.
Note for a finite measure space $E$, (Lebesgue measure), there is a nesting of $L^p(E)$ spaces. We have the following relationship derived from Holder's inequality: for $1\leq p<q\leq \infty$ $\|f\|_{L^p}\leq \mu(E)^{1/p-1/q}\|f\|_{L^q}$
So we have $\|f\|_Y \leq C\|f\|_X$ where $C=\mu(E)^{1/p-1/q}$. So the inclusion map is continuous, and so $X$ is continuously embedded into $Y$.
However, $Y$ is not continuously embedded into $X$.
we can show the inverse inclusion map $\iota^{-1}(f)\mapsto f$ in general is not bounded by taking $f\in L^p(a,b)\setminus L^q(a,b)$, and noting the norm in $L^p$ is finite but the norm in $L^q$ is infinite, showing the inverse/backwards inclusion map is not continuous. I could use the explicit counterexample: $x^{-1/q}{\chi(a,b)}$ to show this.
Consider $(a,b)=(0,1)$, $r\in (p,q)$ and $$ f_\varepsilon(x)=(x+\varepsilon)^{-\frac{1}{r}}. $$ Then $$ \int_0^1 |\,f_\varepsilon|^p<\int_0^1 x^{-p/r}\,dx=\frac{1}{1-p/r}=\frac{r}{r-p} $$ and hence $$ \|\,f_\varepsilon\|_p<\left(\frac{r}{r-p}\right)^{1/p}. $$ Meanwhile $$ \lim_{\varepsilon\to 0}\int_0^1 |\,f_\varepsilon|^q=\lim_{\varepsilon\to 0}\int_0^1 (x+\varepsilon)^{-\frac{q}{r}}\,dx=\lim_{\varepsilon\to 0}\frac{1}{q/r-1}\left(\varepsilon^{-q/r}-(1+\varepsilon)^{-q/r}\right)=\infty. $$