Prove that any linear function \(L: \mathbb{R}^2 \to \mathbb{R}^2\) is continuous.
Solution
1. Let\[ L(x,y)=(ax+by,cx+dy) \]for some \(a,b,c,d\in\mathbb R\).
A linear map \(\mathbb R^2\to\mathbb R^2\) is represented in the standard bases by a \(2\times2\) real matrix.
2. There is a constant \(C\ge0\) such that\[ \|L(v)\|\le C\|v\| \qquad(v\in\mathbb R^2). \]
For \(v=(x,y)\), Cauchy–Schwarz gives \[ |ax+by|^2\le(a^2+b^2)(x^2+y^2), \] and \[ |cx+dy|^2\le(c^2+d^2)(x^2+y^2). \] Hence \[ \|L(v)\|^2 \le(a^2+b^2+c^2+d^2)\|v\|^2. \] Thus the claim holds with \[ C=\sqrt{a^2+b^2+c^2+d^2}. \]
3. The map \(L\) is continuous at every \(v_0\in\mathbb R^2\).
If \(C=0\), then \(L=0\) and continuity is immediate. Suppose \(C>0\). Given \(\varepsilon>0\), choose \(\delta=\varepsilon/C\). If \(\|v-v_0\|<\delta\), then by linearity and 2, \[ \|L(v)-L(v_0)\|=\|L(v-v_0)\|\le C\|v-v_0\|<\varepsilon. \]
4. Therefore every linear map \(\mathbb R^2\to\mathbb R^2\) is continuous.