# Does convergence in $W^{1,1}([0,1])$ imply uniform convergence?

Let $I=[0,1]$. Let $f_n \in W^{1,1}(I)$ be continuous and suppose that $f_n$ converge in $W^{1,1}(I)$ to a smooth function $f$. Does there exist a uniformly convergent subsequence of $f_n$?

(There exist an a.e. convergent subsequence). Sobolev embedding/Morrey's inequality do not help here, since $p=1$ equals the dimension $d=1$. (We don't have $p>d$).

If it helps, we can assume the $f_n$ are smooth.

## 1 Answer

Yes. Indeed, the whole sequence is uniformly convergent.

To see this, first note for a continuous $f \in W^{1,1}$ that $$\min_x |f(x)| \leq \int_0^1 |f| dx .$$ Let's say $\min_x |f(x)| = |f(x_0)|$. Now, if $x \in [0,1]$ is arbitrary, then the fundamental theorem of calculus shows $$|f(x)| \leq |f(x) - f(x_0)| + \min_y |f(y)| \leq \int_{0}^1 |f'(t)| dt + \min_y |f(y)| \leq \|f\|_{W^{1,1}}.$$

Therefore, If we apply this estimate to $f_n -f$ instead of $f$, we see $$\|f_n -f\|_{L^\infty}\leq \|f_n -f\|_{W^{1,1}} \to 0$$ as $n \to \infty$.