Yes, it follows from the classical result of S. Bernstein :

For an arbitrary scheme $X=\cup_{n} A_{n}$ of points in $[-1,1]$, there exist a continuous function $f$ and a point $x_{0}$ in $[-1,1]$ such that $$ { \limsup _ { n \rightarrow \infty } } \left| L _ { n } \left( f, X , x _ { 0 } \right) \right| = \infty. $$ Let $m=\min_{[-1,1]}f$ and $M=\max_{[-1,1]}f$. If there is a subsequence $n_{k}$ with $$ { \lim_ { n_{k} \rightarrow \infty } } ~L _ { n_{k} } \left( f, X , x _ { 0 } \right) = -\infty,$$ just consider the positive function $g(x)=f(x)-m+1>0$. If there is a subsequence $n_{k}$ with $$ { \lim_ { n_{k} \rightarrow \infty } } ~L _ { n_{k} } \left( f, X , x _ { 0 } \right) = \infty,$$ consider the positive function $g(x)=-f(x)+M+1>0$.

Yes, it follows from the following result of S. Bernstein (Quelques remarques sur l'interpolation, Math. Ann. 79 (1918), 1-12):

For an arbitrary scheme $X=\cup_{n} A_{n}$ of points in $[-1,1]$, there exist a continuous function $f$ and a point $x_{0}$ in $[-1,1]$ such that $$ { \limsup _ { n \rightarrow \infty } } \left| L _ { n } \left( f, X , x _ { 0 } \right) \right| = \infty. $$

If there is a subsequence $n_{k}$ with $$ { \lim_ { n_{k} \rightarrow \infty } } ~L _ { n_{k} } \left( f, X , x _ { 0 } \right) = -\infty,$$ consider the positive function $g(x)=f(x)-m+1>0$, where $m=\min_{[-1,1]}f$.

If there is a subsequence $n_{k}$ with $$ { \lim_ { n_{k} \rightarrow \infty } } ~L _ { n_{k} } \left( f, X , x _ { 0 } \right) = \infty,$$ consider the positive function $g(x)=-f(x)+M+1>0$, where $M=\max_{[-1,1]}f$.