# Can the infimum becomes minimum in the following definition?

Let $\mathscr{X}$ be a n.v.s., $M$ be a subspace of $\mathscr{X}$, we know that the definition of norm in $\mathscr{X} /M$ inherited by $\mathscr{X}$ is $$\Vert x+M\Vert_0 =\inf_{y\in M}\Vert x-y\Vert .$$ Can the infimum becomes minimum in the definition? Or under a stronger assumption that $\mathscr{X}$ is a Banach space, or $M$ is closed, can the definition be strengthened to minimum? If not, what is the counterexample?

• You'll need closedness to ensure the infimum is attained. A counterexample, is simple, take $x \in \overline{M} \setminus M$. If you have closedness then I think you're good by some convexity argument. – Ian Apr 13 '18 at 23:00
• Thanks. So it is convenient for us to define the orthogonal decomposition in only an n.v.s. ,can we? – TheWildCat Apr 14 '18 at 3:05
• You need a Hilbert space to do projections. – Ian Apr 14 '18 at 4:47
• If the infimum becomes minimum, we can always define the "orthogonal vector" of the subspace $M$ to be the vector which satisfies $\Vert x+M\Vert_0 =\Vert x\Vert$, and we can similarly directly define an "orthogonal decomposition" without the structure of Hilbert space. – TheWildCat Apr 14 '18 at 12:40
• You can have a look at this post and the questions linked there: On the norm of a quotient of a Banach space. – Martin Sleziak Dec 8 '18 at 11:42