Let $T$ be a bounded linear operator on a Banach space $E$, and let $(T_n)$ be a sequence of bounded linear operators on $E$ such that $$\| (T - T_n)x \| \to 0 \ \ \text{ as } n \to \infty$$ for all $x \in E$. Is it true that for any polynomial $p$ with complex coefficients we have that $$\|(p(T) - p(T_n))x\| \to 0 \ \ \text{ as } n \to \infty$$ for all $x \in E$.
If it is true, can we replace $p$ with any other function for which $f(T)$ makes sense?
Thank you for your help.
We can prove by induction on $k$ that $\lVert _nT^k(x)-T(x)\rVert\to 0$. Indeed, $$T_n^kx-T^kx=T_n(T_n^{k-1}x-T^{k-1}x)+(T_n-T)T^{k-1}x,$$ hence $$\Vert T_n^kx-T^kx\rVert\leqslant\sup_j\lVert T_j\rVert\cdot \lVert T_n^{k-1}x-T^{k-1}x\rVert+\lVert T_n(T^{k-1}x)-T(T^{k-1}x))\rVert.$$ Using the uniform boundedness principle, we obtain that $\sup_j\lVert T_j\rVert$ is finite and the conclusion follows.
Then we can use an approximation argument to show that $\lVert f(T_n)(x)-f(T)(x)\rVert\to 0$ for any continuous function $f$.