Prove that the operator $T:X\to Y$ satisfies $\|T^n\|\leqslant \|T\|^n$.

Question

Let $X$ and $Y$ be normed spaces and $T:X\to Y$ a linear operator. How do I prove this for $T$? Thanks.

Answer 1

Note that you can't compose $T$ with itself, unless $X=Y$.

Let $A,B \in \mathscr{B}(X)$, i.e. bounded linear operators. Then, $\forall x \in X$, we have that $$\lVert Ax \rVert \leqslant \lVert A \rVert \lVert x \rVert$$ And also $$\lVert A Bx \rVert \leqslant \lVert A \rVert \lVert Bx \rVert \leqslant \lVert A \rVert \lVert B\rVert \lVert x \rVert$$ Which means that $$\lVert AB \rVert \leqslant \lVert A \rVert \lVert B \rVert$$ i.e. the operator norm is submultiplicative. Now you can use induction to prove what you want.