Question

If \(c \in {V_n}\), show that the function \(f\) given by \(f(x)\), \(f(x) = c.x\) is continuous on \({R^n}\).

Short answer

\(f(x) = c.x\) is continuous on \({R^n}\).

Step-by-step solution

Finding \(\delta  > 0\)

Let us assume that \(\varepsilon > 0\)

We need to find \(\delta > 0\) such that \(\left| {f(x) - f(a)} \right| < \varepsilon \) whenever \(\left| {x - a} \right| < \delta \) or \(\left| {cx - ca} \right| < \varepsilon \)whenever \(\left| {x - a} \right| < \delta \)

But \(\left| {cx - ca} \right| = \left| {c.(x.a)} \right|\) and \(\left| {c.(x - a)} \right| \le \left| c \right|\left| {x - a} \right|\)

Assuming \(\varepsilon  > 0\):

Let \(\varepsilon > 0\) be given and set \(\delta = \frac{\varepsilon }{{\left| c \right|}}\)

Now, whenever \(0 < \left| {ax - a} \right| < \delta \)

\(\left| {f(x) - f(a)} \right| = \left| {cx - ca} \right| \le \left| c \right|\left| {x - a} \right| < \left| c \right|\delta \)

\( = \left| c \right|(\varepsilon \left| {\left| c \right|} \right|)\)

\( = \varepsilon \)

Hence, \(f\)is continuous on \({R^n}\).