# How to show $\lim\limits_{t\to 0}\frac{|f+tg|^{p}-|f|^{p}}{t}=pfg|f|^{p-2}$

Could anyone help to compute this limit? Thank you!

This is part of a proof in Analysis by E. Lieb. Let $f$ and $g$ be real numbers and $p>1$, show the following limit: $$\lim\limits_{t\to 0}\frac{|f+tg|^{p}-|f|^{p}}{t}=pfg|f|^{p-2}$$

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If $g=0$, both sides are constant $0$.

If $g\neq 0$ and $f=0$, we have $$\lim_{t\to 0}\frac{|tg|^p}{t} = |g|^p\lim_{t\to 0}\frac{|t|^p}{t} = 0,$$ since $p\gt 1$ (it's the limit of $|g|t^{p-1}$ as $t\to 0^+$, which is $0$, and of $-|g|t^{p-1}$ as $t\to 0^-$, which is also $0$).

If $g\neq 0$ and $f\neq 0$, say $f\gt 0$. Using L'Hopital's Rule you have \begin{align*} \lim_{t\to 0}\frac{|f+tg|^p - |f|^p}{t} &= \lim_{t\to 0}\frac{(f+tg)^p - f^p}{t}\\ &= \lim_{t\to 0}\frac{p(f+tg)^{p-1}g - 0}{1} = pf^{p-1}g = pfgf^{p-2} = pfg|f|^{p-2}. \end{align*}

If $f\lt 0$, then \begin{align*} \lim_{t\to 0}\frac{|f+tg|^p - |f|^p}{t} &= \lim_{t\to 0}\frac{(-f-tg)^p - (-f)^p}{t}\\ &= \lim_{t\to 0}\frac{p(-f-tg)^{p-1}(-g) - 0}{1} \\ &= p(-f)^{p-1}(-g)\\ &= -pg|f|^{p-1}\\ &= p(-|f|)g|f|^{p-2}\\ &= pfg|f|^{p-2}. \end{align*}

In any of the four cases, we have equality.

(As to getting rid of the absolute value signs in $|f+tg|$, so long as $f\neq 0$, $f+tg$ has the same sign as $f$ for sufficiently small $t$).

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The given limit is easily seen to be the (definition of the) derivative of the function $$t \mapsto |f+gt|^p$$ at $t = 0$. We can calculate this derivative using the chain rule. This comes out to be $$p |f+gt|^{p-1} \times \frac{f+gt}{|f+gt|} \times g.$$ Here, the first factor is the derivative of the function $z \mapsto z^p$ at $|f+gt|$. The second factor is the derivative of the absolute value function at $f+gt$. Upon substituting $t=0$, we get the claimed result.

To make the above proof watertight, observe that the absolute value function is differentiable only when the argument is nonzero; so in the above, we have to assume that $f \neq 0$. One would then need to complete the proof by discussing the case $f=0$ separately, as in @Arturo Magidin's answer.

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