What can we say about the bound of a function given its $L^1$ bounds of its partial derivatives

Question

Suppose $f \in C_0 ^\infty (\mathbb R^d ) $ be the space of smooth function which vanishes at infinity and we know that $\| \partial_i f\|_{L^1 (\mathbb R^d ) } \le M_i $ for all $1\le i\le d$. Can we say something about $\|f\|_{L^{\infty}} $ in terms of $M_i$?

Answer 1

No, there is no such bound. In fact, we can construct a sequence $f_n$ with $\|\partial_if_n\|_{L^1(\mathbb R^d)}\leq M$, but $\|f\|_{L^{\infty}(\mathbb R^d)}\to \infty$.

To do this, consider any nonzero $f\in C_0(\mathbb R^d)$ with $\|\partial_if\|_{L^1(\mathbb R^d)}\leq M$, and set $f_n(x)=n^{d-1}f(nx)$. Then $f_n\in C_0(\mathbb R^d)$, $$\|\partial_if_n\|_{L^1(\mathbb R^d)}=n^{d-1}\int_{\mathbb R^d}|\partial_i(f(nx))|\,dx=n^d\int_{\mathbb R^d}|\partial_if(nx)|\,dx=\int_{\mathbb R^d}|\partial_if(y)|\,dy=\|\partial_iφ\|_{L^1(\mathbb R^d)},$$ which is bounded by $M$, but $$\|f_n\|_{\infty}=n^{d-1}\|f\|_{\infty}\to\infty.$$