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Given a linear maps, $A, B : \mathbb{R}^n \to \mathbb{R}^m$, and a norm $\|A\| = \sup_{|x| \le 1} |Ax|$, I want to show that given $\varepsilon > 0$, there exists $\delta > 0$ such that $\|A-B\| < \delta$ implies $|Ax - Bx| < \varepsilon$.

I am not actually sure if that is true or not, but it feels like it should be true.

I found that if I restrict myself to the ball where $|x| < r$, I can get $$|Ax - Bx| = |(A - B)x| \le \|A-B\||x| < \delta r.$$ Thus given $\varepsilon > 0$, I can let $\delta = \varepsilon / r$.

However, can I extend this argument to all of $\mathbb{R}^n$?

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    You need the norm constraint -- doubling the norm of $x$ will double $\lVert A x - B x \rVert$. Doing enough doubling will make it exceed any $\epsilon$. – 2017-02-22
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    Been away from math stuff for a bit, heh. What is the norm constraint? – 2017-02-22
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    It might be easiest to understand batman's comment for $n = m = 1$. Suppose $A$ is multiplication by $3$. How close must $B$ be to $3$ to ensure that $ | (A-B) x | < .01$? If you pick $B = 2.99$, then this works fine...: $| (3 - 2.99) x | = |.01 x| < .01$ if $|x| < 1$. But if you pick $x = 1000$, then $.01 x = 10$ and you're in trouble. – 2017-02-22
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    The "norm constraint" is your assumption that $|x| < r$. – 2017-02-22
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    Oh duh. Hmm, for some reason I thought this was doable. If I consider the matrix form of $A$, and I perturb each of the components $a_{ij}$ by some small amount, such that I get a diagonalizable matrix, $B$, and it does not differ much in output, i.e. $|Ax - Bx| < \epsilon$. Would that be limited to the same norm constraint, or can that be done on all of $\mathbb{R}^n$? I might be remembering this incorrectly. – 2017-02-22
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    Diagonalization doesn't help -- so long as $A \neq B$, you can find an $x$ such that $(A-B) x \neq 0$ so $\lVert A x - B x \rVert \neq 0$. Then, $\lVert (A - B) (2^n x) \rVert = 2^n \lVert A x - B x \rVert$. Now, you can pick $n$ as large as you want in order to violate the bound. – 2017-02-22
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    Why do you think the definition of the norm of $A$ has the constraint in it? – 2017-02-22

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There is no $\varepsilon > 0$ such that $\lVert Ax - Bx\rVert < \varepsilon$ for every $x\in\mathbb R^n$, unless $A = B$, because if $y\in\mathbb R^n$ is such that $$\lVert Ay - By\rVert=\Delta >0,$$ then $$\lVert A(m y /\Delta)-B(m y /\Delta)\rVert=|m|$$ can be made arbitrarily large by appropriately choosing $m\in\mathbb R$. What you can do is show this for every $x$ of unit norm or, equivalently, show that there is some $\varepsilon >0$ such that $\lVert Ax - Bx\rVert\leq\varepsilon \lVert x\rVert$ for every $x\in\mathbb R^n$.