# derivation of an inequality

I'm reading a text and sometimes it skips steps in its derivations that I don't always understand. An example follows.

Suppose we have, $f(y) \ge f(x) + \frac{1}{1-\alpha}\left[f(x + (1-\alpha)(y-x))-f(x)\right]$.

The next line in the text says that, as $\alpha \longrightarrow 1$, the above relation becomes $f(y) \ge f(x) + \langle f^{\prime}(x), x-y)\rangle$ where $\langle\ \rangle$ denotes the dot product.

Can someone explain why that is ? To me, the term in [ ], looks like it goes to $f^{\prime}(x)$ not $\langle f^{\prime}(x), y-x)\rangle$. Note that I did find a typo earlier in the book so there may be some typo here ? Thanks a lot.

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Looks like you received several excellent answers to your previous questions. Please read this page to see how to accept answers: meta.stackexchange.com/a/5235 – Ayman Hourieh Jan 13 '13 at 19:46
Accepting an answer rewards posters for solving your problem and informs others that your issue is resolved. This is why it's a good idea to do so when you receive an answer that you're satisfied with. I don't think the check mark ever expires. Perhaps you were looking at someone else's question by mistake? – Ayman Hourieh Jan 13 '13 at 22:16

## 1 Answer

Fix $x$ and $y$ (where $x \ne y$). Let $\delta = (1 - \alpha)(y-x)$. As $\alpha \to 1$, we have $\delta \to 0$. Thus: $$\lim_{\alpha \to 1} \frac{f\left(x + (1-\alpha)(y-x)\right) - f(x)}{(1-\alpha)} = (y -x)\lim_{\delta \to 0} \frac{f(x + \delta) - f(x)}{\delta} = (y - x) f'(x)$$

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Thank you very much for your beautiful answer. I understand it now and it's appreciated. – mark leeds Jan 13 '13 at 21:58
@markleeds Glad to know it helped! – Ayman Hourieh Jan 13 '13 at 22:13