Let $d \in \mathbb{N}$ and $p \in [1,\infty)$. Does there exist $C>0$ such that for every $f \in W^{1,p}([0,1],\mathbb{R}^d)$ we have $$\|f\|_{L^\infty}\leq C\|f\|_{L^p}^{1-\frac{1}{p}}\|f\|_{W^{1,p}}^{\frac{1}{p}}?$$ My Motivation is the following: For $m \in \mathbb{N}$ I have an approximation operator $Q^h$ depending on the meshwidth $h>0$ such that for every $l\leq m$ there exist $C>0$ such that for every $f \in W^{m,p}$ we have $$\|f-Q^hf\|_{W^{l,p}}\leq Ch^{m-l}\|f\|_{W^{m,p}}.$$ Now I want to show that $\|f-Q^hf\|_{W^{m-1,\infty}} \rightarrow 0$. The statement above would imply that $$\|f-Q^hf\|_{W^{m-1,\infty}}\leq C h^{1-\frac{1}{p}}\|f\|_{W^{m,p}}.$$

Let $p \in [1,\infty)$. Does there exist $C>0$ such that for every $f \in W^{1,p}([0,1],\mathbb{R})$ we have $$\|f\|_{L^\infty}\leq C\|f\|_{L^p}^{1-\frac{1}{p}}\|f\|_{W^{1,p}}^{\frac{1}{p}}?$$ My Motivation is the following: For $m \in \mathbb{N}$ I have an approximation operator $Q^h$ depending on the meshwidth $h>0$ such that for every $l\leq m$ there exist $C>0$ such that for every $f \in W^{m,p}$ we have $$\|f-Q^hf\|_{W^{l,p}}\leq Ch^{m-l}\|f\|_{W^{m,p}}.$$ Now I want to show that $\|f-Q^hf\|_{W^{m-1,\infty}} \rightarrow 0$. The statement above would imply that $$\|f-Q^hf\|_{W^{m-1,\infty}}\leq C h^{1-\frac{1}{p}}\|f\|_{W^{m,p}}.$$