Sequence in $C[0,1]$ with no convergent subsequence

Solution 1:

Since $f_n$ converges to $f$ pointwise, all subsequences $f_{n_k}$ also converge pointwise to $f$. So $f_{n_k}$ can not converge uniformly to continuous $g$, because uniform convergence implies pointwise convergence as well, and $f$ is not continuous.

Solution 2:

This is (more-or-less) a modification of the hint from t.b's comment.

It suffices to show that any subsequence of $(f_n)$ is not Cauchy. (Since every convergent sequence is a Cauchy sequence.)

To do this, notice that $$\lVert f_{2n}-f_n \rVert = \sup_{x\in[0,1]} (x^n - x^{2n}) = \sup_{x\in[0,1]} (x^n - (x^{n})^2) = \sup_{t\in[0,1]} (t-t^2)=\frac14.$$ (It is easy to find maximum of the quadratic function $f(t)=t-t^2=t(1-t)$.)

In addition, if we use the monotonicity of the sequence $(f_n)$, we see that $$k\ge 2n \qquad \Rightarrow \qquad \lVert f_{k}-f_n \rVert \ge \frac14.$$

Now, if we have any subsequence of $(f_n)$ then the above estimate shows that this subsequence is not Cauchy. For any given $k_0$ we can find $k'>k_0$ such that $n_{k'}>2n_{k_0}$ and $\lVert f_{n_{k'}}-f_{n_{k_0}} \rVert \ge \frac 14$.