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I'm devastatingly incompetent at linear algebra and multivariable calculus. I just cannot understand it at all. Here's the easiest problem from my homework, and my attempt at solving it, and where I am stuck. Hopefully if I find at least one point of illumination in this problem, I may be able to find more light in this cold dark dungeon.

The Problem:

Suppose that $f:R^n\to R^m$ is linear. Show that $Df(x)$ exists for each $x\in R^n$ and $Df(x)=f(x)$

My attempt:

Take any $x\in R^n$. Consider $$\lim_{h\to 0}\|f(x+h)-f(x)-Th\|/\|h\| = \lim_{h\to 0}\|f(x)+f(h)-f(x)-Th\|/\|h\| = \lim_{h\to 0}\|f(h)-Th\|/\|h\|$$

If there exists a linear map T such that this limit equals 0, then f is differentiable at x and $Df(x)=T$.

My issues:

I don't know where to go from there. If I assumed that $Th=Df(h)=f(h)$, I know that $$\lim_{h\to 0}\|f(h)-f(h)\|/\|h\|=0$$, but I don't know how to prove it the other way around.

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    $\begingroup$ You have $Th$, not $Tx$, so the limit is $$\lim_{h\to 0}|f(h)-f(h)|/|h|$$ Hopefully it is now clear that this is zero :) $\endgroup$ – Zev Chonoles Feb 3 '14 at 5:20
  • $\begingroup$ do you know examples of such $f$ 's? $\endgroup$ – janmarqz Feb 3 '14 at 5:21
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    $\begingroup$ Use \| for the double-barred norm. $\endgroup$ – user61527 Feb 3 '14 at 5:23
  • $\begingroup$ What do you mean by proving it "the other way around"? $\endgroup$ – Santiago Canez Feb 3 '14 at 5:28
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    $\begingroup$ Actually, I think Zev's comment helped clarify things though. Clearly Tx=f(x) is a linear map where the limit equals 0, so f is differentiable and Df(x)=Tx=f(x). It is a basic property that derivatives are unique. I don't know how to mark a question answered if all responses are comments. What do I do? $\endgroup$ – Jeff Feb 3 '14 at 5:32
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Consider

$$\lim_{h\to 0}{\|f(x+h)-f(x)-Th\| \over \|h\|}$$

If there exists a linear map $T$ such that this limit equals $0$, then $f$ is differentiable at $x$ and $Df(x)=T$.

Note that by assumption $f$ is linear. Hence $f(x+h)=f(x) + f(h)$. Now take $T=f$. Then $$\lim_{h\to 0}{\|f(x+h)-f(x)-Th\| \over \|h\|} = \lim_{h\to 0}{\|f(x)+f(h)-f(x)-f(h)\| \over \|h\|} = \lim_{h \to 0} {0 \over \|h\|}$$

Certainly, $\lim_{h \to 0} {0 \over \|h\|}=0$ hence $f$ is differentiable at $x$. Since $x$ was arbitrary, $f$ is differentiable everywhere.

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