How to prove that following supremum is attained by the function.

Question

Given $T$ is a bounded linear operator and we are working in Hilbert space. If i have $$\alpha(\delta, y^{\delta}) = \sup \left\{\alpha>0, \ | \ \left\lVert Tx_{\alpha}-y^{\delta}\right\rVert \leq c \delta\right\}$$ where $c >1$. If i know that $$\alpha \to \left\lVert Tx_{\alpha}-y^{\delta}\right\rVert$$ is continuous from left. Then how we can say that supremum in above is attained.

Answer 1

I assume that $\delta$ is fixed. Define $f(\alpha) :=\left\lVert Tx_{\alpha}^{\delta}-y^{\delta}\right\rVert$. If the set $I:=\left\{ \alpha\gt 0\mid f(\alpha)\leqslant c\delta\right\}$ is empty, its supremum is $0$. Otherwise, we can construct a sequence $(\alpha_n)_n$ such that $\alpha_n\in I$ for each $n$ and $\alpha_n\gt \sup I- 1/n$. We can extract an increasing subsequence (that we will denote in the same way). Then $\alpha_n\to \sup I=:\alpha$ and since $f(\alpha_n)\leqslant c\delta$ for each $n$, then by left-continuity, we have $f\left(\alpha\right) \leqslant c\delta$ hence $\alpha$ belongs to $I$.