Suppose $X$ is a Banach space, and $\{x_n \}\subset X$. Does it then hold that $\lim \|y-x_n\|=\|y-\lim x_n \| $ ?
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If $(x_n)$ converges to $x$, then yes; since $y-x_n$ then converges to $y-x$. – David Mitra Apr 07 '14 at 21:30
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You are wondering whether the norm is continuous with respect to itself. The answer is trivially affirmative. – Giuseppe Negro Apr 07 '14 at 21:32
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First let's recall what it means to say $\displaystyle\lim_{n\to\infty} x_n = x$ in a Banach space. It means $\displaystyle\lim_{n\to\infty}\|x_n-x\|=0$. The first limit is a limit of a sequence of members of the Banach space; the second is a limit of a sequence of non-negative numbers.
Note that $\|A\|=\|(A-B)+B\|\le \|A-B\|+\|B\|$, so that $$\|A\|-\|B\|\le\|A-B\|,$$ and in the same way show that $$\|B\|-\|A\|\le\|A-B\|,$$ so we have $$ |\, \|A\|-\|B\| \,|\le \|A-B\|. $$ Hence $$ |\,\|y-x_n\|-\|y-x\|\,| \le \|(y-x_n)-(y-x)\| = \|x_n-x\| \tag 1 $$ Consequently, if the rightmost term in $(1)$ goes to $0$, then so does the leftmost term.