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I seek something of the form $\|f-g\|_p \leq |\|f\|_p-\|g\|_p|$.

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2 Answers 2

up vote 5 down vote accepted

You get exactly the opposite inequality. From the standard Minkowski inequality you get $\|f\|_p = \|(f-g)+g\|_p \le \|f-g\|_p + \|g\|_p$, so $\|f-g\|_p \ge \|f\|_p - \|g\|_p$. Also, $\|f-g\|_p = \|g-f\|_p \ge \|g\|_p - \|f\|_p$ by reversing the roles of $f$ and $g$. Combining both inequalities gives $\| f-g \|_p \ge | \|f\|_p - \|g\|_p|$

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We know that $\lVert f_1+g_1\rVert_p\leq\lVert f_1\rVert_p+\lVert g_1\rVert_p$ for (appropriate) $f_1,g_1$. In particular, if we take $f_1=f-g$ and $g_1=g$ (for appropriate $f,g$), then we have $$\lVert f\rVert_p\leq\lVert f-g\rVert_p+\lVert g\rVert_p,$$ and so $$\lVert f\rVert_p-\lVert g\rVert_p\leq\lVert f-g\rVert_p.$$ On the other hand (using a similar approach), we have $$\lVert g\rVert_p-\lVert f\rVert_p\leq\lVert g-f\rVert_p=\lVert f-g\rVert_p.$$ Thus, it follows that $$\bigl|\lVert f\rVert_p-\lVert g\rVert_p\bigr|\leq\lVert f-g\rVert_p.\tag{#}$$ Your inequality, then, is precisely backwards.

For a more intuitive view, the left-hand side of $(\#)$ has some guaranteed cancellation, while the right-hand side may actually build upon itself. As an example, consider the case that $g=-f$ with $\lVert f\rVert_p>0$.

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