I can recall that from here about 'the sequence $(a_n)$ has no converging subsequence if and only if $|a_n|$ tends to infinity'. Can I extend this lemma into the sequence of functions and replaces the modulus function with a norm function? Would the result still hold?
Also are there other strategies to show a sequence of functions do not have a converging subsequence?
On
I'm presuming that the question you're asking above is asking "can I replace the absolute value in the above statement with a modulus without losing the result?" Well the main problem with your question is that modulus takes two arguments, absolute value only takes one. And with any finite modulus (say, $M$), then your new sequence becomes $(a_n \mod M)$, which is obviously bounded and thus convergent by Bolzano-Weierstrass: $$ (a_n\mod M)\subseteq[0,M)\subset [0,\infty) $$ $$ \Downarrow $$ $$ a_{n_k}\in (a_n\mod M ), a_{n_k}\rightarrow a \text{ for some }a\in[0,M] $$ As for the second half of this question, a very useful theorem for you would be the Bolzano-Weierstrass Theorem, which states: any bounded sequence of real numbers has a convergent subsequence.
The Bolzano-Weierstrass theorem tells us that any bounded sequence in $\mathbb{R}^n$ has a convergent subsequence. This does not hold for metric spaces in general, and fails in most function spaces.