For a linear operator $R \in B(X,Y)$ and some $k \in \mathbb{R}_{>0}$, do we have $$(\forall x \in X:\lVert R(x) \rVert \leq k \lVert x \rVert) \Rightarrow \lVert R \rVert \leq k $$?
I'd like to use this fact to prove the triangle inequality for norms of linear operators $S, T \in B(X,Y)$, of which I know that
$$\forall x \in X:\lVert (S+T)(x)\rVert \leq (\lVert S \rVert + \rVert T \lVert) \lVert x\rVert $$
Note that $$\|R\|=\sup\{\|R(x)\|:x\in X,\|x\|=1\}.$$ Indeed, write $p(R)=\sup\{\|R(x)\|:x\in X,\|x\|=1\}$. Since $$\{\|R(x)\|:x\in X,\|x\|=1\}\subset\{\|R(x)\|:x\in X,\|x\|\leq1\},$$ we have $p(R)\leq\|R\|$. On the other hand, if $0<\|x\|\leq1$, then $$\|R(x)\|\leq\frac{1}{\|x\|}\|R(x)\|=\left\|R\left(\frac{x}{\|x\|}\right)\right\|\leq p(R).$$ Taking the supremum over all such $x$ yields $\|R\|\leq p(R)$, and thus $\|R\|=p(R)$.
If now $x\in X$ and $\|x\|=1$, then by assumption $\|R(x)\|\leq k\|x\|=k$. Thus $k$ is an upper bound of $\{\|R(x)\|:x\in X,\|x\|=1\}$, whence $\|R\|\leq k$.