The tittle of this question is the main question.
Why it is necessary to not consider these kind of functions in $\Lambda$. My teacher told me that if one consider these functions to be in the ring of symmetric functions, then the inner product would fail since this kind of functions would have infinitely non-zero homogeneous parts. But I do not see it clearly.
Can anyone explain me why it is necessary to exclude the functions which do not have bounded degree?
Thank you very much.
$$\prod_i (1+x_i) \notin \Lambda.$$
I agree that it is difficult to see how $f:=\prod_i(1+x_i)$ expands into homogenous poylnomials. But it is still easy to see why a contradiction arises.
If we were to expand $f$ into elementary symmetric functions, the result would be the infinite sum $$ f=1+e_1+e_2+\dots $$ This is the extension of the finite equality $$(1+x_1)(1+x_2)(1+x_3)=1+(x_1+x_2+x_3)+(x_1x_2+x_1x_3+x_2x_3)+x_1x_2x_3.$$ Now, applying the $\omega$ involution to both sides, $$ \omega(f)=1+h_1+h_2+\dots $$ Next, expand $\omega(f)$ into the $m$ basis. The result is $$ \omega(f)=\sum_{\lambda }m_\lambda, $$ since each $m_\lambda$ appears in exactly one $h_n$. Finally, we can see that (using the convention $h_{(0)}=m_{(0)}=1$ for convenience) $$ \langle \omega(f),\omega(f)\rangle=\left\langle \sum_{i=0}^\infty h_{(i)},\sum_\lambda m_\lambda\right\rangle=\sum_{i=0}^\infty \langle h_{(i)},m_{(i)}\rangle = \sum_{i=0}^\infty 1=\infty $$ We do not want infinite values of the inner product. It may not seem like problem, but if $\infty$ can arise, then so can $\infty-\infty$, which is a problem. For example, computing the inner product of $\sum_{i=1}^\infty (-1)^i h_{(i)}$ and $\sum_{\lambda}m_\lambda$ would be $1-1+1-1+\dots$, which is undefinable.