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I have to prove that a sequence $A(\varepsilon)$ of operators between Hilbert spaces $A(\varepsilon):H_1\to H_2$ converges, when $\varepsilon\to 0^+$, to an operator $B:H_1\to H_2$ in the uniform norm of the space $B(H_1,H_2)$ (the space of bounded operators from $H_1\to H_2$). So I have to prove that $$\Vert A(\varepsilon)-B\Vert_{B(H_1,H_2)}\to 0,\,\,\,\varepsilon\to 0^+$$ where $$\Vert A\Vert_{B(H_1,H_2)}:=\sup_{f\in D(A)}\frac{\Vert Af\Vert_{H_1}}{\Vert f\Vert_{H_2}}$$. I've found that for every $f\in H_1$ $$\Vert A(\varepsilon)f-Bf\Vert_{H_1}\leq C(\varepsilon)\Vert f\Vert_{H_2}$$ with the quantity $C(\varepsilon)\to 0$ when $\varepsilon\to 0^+$. Can I conclude?

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1 Answer 1

Since $$\Vert A(\varepsilon)f-Bf\Vert_{H_1}\leq C(\varepsilon)\Vert f\Vert_{H_2}$$ implies $$\frac{\Vert A(\varepsilon)f-Bf\Vert_{H_1}}{\Vert f\Vert_{H_2}}\leq C(\varepsilon),$$ you have $$\Vert A(\varepsilon) - B\Vert_{B(H_1,H_2)}:=\sup_{f\in D(A)}\frac{\Vert A(\varepsilon)f - Bf\Vert_{H_1}}{\Vert f\Vert_{H_2}} \le C(\varepsilon).$$

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So I have the desired convergence! – Mary Apr 15 '13 at 9:44

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