# Is there a sequence $(f_n)_n \subset C_b(\mathbb{R})$ with $f_n \to f \in C_b(\mathbb{R})$ pointwise but $f_n(1/n) \not \to f(0)$?

Can we construct a sequence of functions in the space $$C_b(\mathbb{R})$$ (continuous and bounded functions) such that

$$f_n(x) \to f(x) \ \ \forall x \in \mathbb{R}$$

where $$f \in C_b(\mathbb{R})$$ with the additional property that $$f_n(1/n) \not \to f(0)$$

Yes, there is. Take $$f_n$$ to be a function whose graph is a triangle based at the vertices $$(0,0),(0,2/n)$$ and $$(1/n,n)$$, and zero elsewhere. Clearly $$f_n$$ is continuous for each $$n$$, and for every $$x\in\mathbb{R}$$ the sequence $$f_n(x)$$ tends to zero, but $$f_n(1/n)=n$$.

If you want a sequence of uniformly bounded continuous functions, change the vertex $$(1/n,n)$$ to $$(1/n,1)$$, so that $$f_n(1/n)=1$$ for all $$n$$.