$X$ : normed space on $K=\mathbb R, \mathbb C$ with norm $\|\cdot \|$
$L \subsetneq X$ : closed linear subspace.
$a\in X\setminus L.$
Then, define the distance between $a$ and $L$ by $d(a,L)=\inf\{ \|a-l\| \mid l\in L\}.$
Prove $d(a,L)=\inf\{ \|a-l\| \mid l\in L,\ l\neq 0 \}.$
$\leqq$ is obvious since $\{ \|a-l\| \mid l\in L\}\supset \{ \|a-l\| \mid l\in L,\ l\neq 0 \}$, so I have to show $d(a,L)\geqq\inf\{ \|a-l\| \mid l\in L,\ l\neq 0 \}$.
Pick $l\in L$ arbitrarily.
It suffices to show $\|a-l\| \geqq \inf\{ \|a-l\| \mid l\in L,\ l\neq 0 \}.$
If $l\neq 0$, then $\|a-l \|\geqq \inf\{ \|a-l\| \mid l\in L,\ l\neq 0 \}$ is obvious.
If $l=0,$ how should I do ?
If $l=0,$ then $\|a-l\|=\|a\|$ and I cannot show $\|a\|\geqq\inf\{ \|a-l\| \mid l\in L,\ l\neq 0 \}$.
This seems not so difficult but I'm in trouble with showing this.
Thanks for any help.
Fix any $b \neq 0$ in $L$ and note that $\|a-\frac 1n b\| \geq \inf \{\|a-l\|: l \in L, l\neq 0\}$. Take limit as $n \to \infty$. to see that $\|a\| \geq \inf \{\|a-l\|: l \in L, l\neq 0\}$.