Let $E,F, G$ be three Banach spaces and let $(v_n)$ be a sequence of continuous linear functions from F to G which converge to $v ,$ and $(u_n)$ be a sequence of linear and continuous functions from $E$ to $F$ which converge to $u.$
Prove that $v_n\circ u_n$ converges to $v\circ u$ in $\mathcal{L}(E,G)$.
I know that $$\|(v_n \circ u_n)-(v\circ u)\|= \sup_{\|x\|_E\leq 1} \|(v_n\circ u_n)(x)-(v\circ u)(x)\|_G,$$
but I don't know how to continue.
Thank you!
Best Answer
Not sure if this is the fastest answer, but it gets the job done. Recall that a linear function $f$ between Banach spaces is continuous if and only if it is bounded, e.g. $\|f(x)\|\leq C_f\|x\|$.
As suggested in the comments, begin with \begin{align*} \|v_n(u_n(x))-v(u(x))\|_G &\le\|v_n(u_n(x))-v_n(u(x))\|_G + \|v_n(u(x))-v(u(x))\|_G \\ &\le\|v_n(u_n(x)-u(x))\|_G + \|v_n(u(x))-v(u(x))\|_G \\ \end{align*}
Since each $v_n$ is bounded, the first term is at most $C_{v_n}\|u_n(x)-u(x)\|_F$.
Let us take supremums on both sides and assume that we can remove the $n$-dependence from the constant $C_{v_n}$; we will justify this assumption at the end of the post. We then obtain: $$\sup_{\|x\|_E\leq 1}\|v_n(u_n(x))-v(u(x))\|_G ~~\le~~ C\sup_{\|x\|_E\leq 1}\|u_n(x)-u(x)\| + \sup_{\|x\|_E\leq 1}\|v_n(u(x))-v(u(x))\|_G.$$
For sufficiently large $n$, the first term may be made strictly less than $\varepsilon/2$, directly from the convergence $u_n\to u$. The second term may also be made at most $\varepsilon/2$ as follows: since $\|x\|_E\leq 1$ we have $\|u(x)\|_F\leq C_u$. Thus, for sufficiently large $n$, $$\sup_{\|x\|_E\leq 1}\|v_n(u(x))-v(u(x))\|_G \leq \sup_{\|y\|_F\leq C_u}\|v_n(y)-v(y)\|_G = C_u \sup_{\|y\|_F\leq 1}\|v_n(y)-v(y)\|_G<\varepsilon/2.$$
Hence, we have bounded $\|v_n\circ u_n - v\circ u\|$ by $\varepsilon$ for any sufficiently large $n$, as desired.
The "technical details" song-and-dance:
By taking the relevant supremum, we can remove the $n$-dependence in this constant as follows: since $v_n\to v$, we have that $\sup_{\|y\|_F\leq 1}\|v_n(y)-v(y)\|_F\leq 1$ for any $K$ and for all sufficiently large $n$, so that $$\sup_{\|y\|_F\leq 1}\|v_n(y)\|\leq \sup_{\|y\|_F\leq 1}\|v_n(y)-v(y)\|_F + \sup_{\|y\|_F\leq 1}\|v(y)\|_F$$ and thus $C_{v_n}=\|v_n\|\leq 1+\|v\|$ for all sufficiently large $n$. Moreover, since $u_n\to u$, we have that $\sup_{\|x\|_E\leq 1}\|u_n(x)-u(x)\|_F\leq 1$ for all sufficiently large $n$. Since the supremum displayed above therefore includes $y=u_n(x)-u(x)$, we have that for sufficiently large $n$, $$\sup_{\|x\|_E\leq 1}\|v_n(u_n(x)-u(x))\|_G \leq (C_v+1)\sup_{\|x\|_E\leq 1}\|u_n(x)-u(x)\|_F.$$ Notice that $C=C_v+1$ is independent of $n$, as desired.