Directional derivative of a function (Munkres)

Question

Let $A\in \mathbb{R^n}$; let $f:A\rightarrow \mathbb R^n.$ Show that if $f'(a,u)$ exists, then $f'(a,cu)$ exists and equals $cf'(a,u).$

This exercise is from Munkres. I suppose $a\in \mathbb R^n$ and $c\in \mathbb R.$ I tried to calculate $f'(a,cu)$ and I got this:

$f'(a,cu)=\lim_{t\to 0}\frac{ f(a+t(cu))-f(a)}{t}=\lim_{t\to 0}\frac{ f(a+(tc)u)-f(a)}{t}$

I took $r=ct.$ Then, $r\to 0$ since $t\to 0.$ Then

$\lim_{t\to 0}\frac{ f(a+(tc)u)-f(a)}{t}=\lim_{r\to 0}\frac{ f(a+ru)-f(a)}{r}$ and the last limit exists since $f'(a,u)$ does.

Is my argument correct? And how can I show the equality? Thanks in advance!

Answer 1

From here

$$f'(a,cu)=\lim_{t\to 0}\frac{ f(a+t(cu))-f(a)}{t}=c \cdot \lim_{ct\to 0}\frac{ f(a+(tc)u)-f(a)}{ct}$$

Answer 2

Let $f'(a;cu) = A$ and $r=tc$. Remind that $|t|<\delta_0$ $\iff$ $|r|<\delta_1$ for proper choices of $\delta_0$ and $\delta_1$.

So given $\epsilon>0$, we have that there exists $\delta_0$ such that $|t|<\delta_0 \implies \left|\frac{f(a+tcu)-f(a)}{t}-A\right|<\epsilon$.

Then there exists $\delta_1>0$ such that $|r|< \delta_1 \implies \left|\frac{f(a+tcu)-f(a)}{t}-A\right|<\epsilon$.

But $$ \left|\frac{f(a+tcu)-f(a)}{t}-A\right| = \left|c\left(\frac{f(a+tcu)-f(a)}{ct}\right)-A\right| = \left|c\left(\frac{f(a+ru)-f(a)}{r}\right)-A\right|. $$

Hence we showed that $$ \lim_{r\to 0}c\left(\frac{f(a+ru)-f(a)}{r}\right) = A $$ But we can take $c$ out of the limit by usual properties of limits, so we showed that $$ c\lim_{r\to 0}\left(\frac{f(a+ru)-f(a)}{r}\right) = cf'(a;u)= A = f'(a;cu) $$ $$ \implies f'(a;cu) = cf'(a;u). $$