$l_2$-norm Inequality

Question

I have $\|a-b\|^2\leq c$. I need to bound $\|a\|^2$ in terms of $\|b\|^2$ and $c$. While there can be many different upper bounds. The article which I am going through has used $\|a\|^2\leq4(\|b\|^2+c)$. While trying to bound it myself, I was not able to come up with such a bound. I was wondering How tight is this bound and How can we prove that this bound holds?

Answer 1

If we are on an inner product space you can use the fact that for any vectors (using Cauchy-Schwarz in the last inequality) $d,e$ we have $$\|d+e\|^2=\|d\|^2+\|e\|^2+2\langle d,e\rangle\le2(\|d\|^2+\|e\|^2).$$ Now set $d=b, e=a-b$. This bound is sharper than yours as $2<4$.

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Also, if we are not on an inner product space, by Jensen (the norm squared is convex) you can proceed as follows: $$\left\|\frac{d+e}2\right\|^2\le\frac{\|d\|^2+\|e\|^2}2$$ i.e. $$\| d+e\|^2\le 2(\|d\|^2+\|e\|^2)$$