Prove $||x+y||=||x||+||y||$ in a Hilbert Space

Question

Consider an infinite dimensional Hilbert Space. Prove $$||x+y||=||x||+||y||$$.

I immediately thought of two identities:

Parallelogram equality: $||x+y||^2+||x-y||^2=2(||x||^2+||y||^2)$

Cauchy-Schwarz equality: $\langle x,y \rangle\leqslant ||x||||y||$

I realized that if $x=\lambda y=\frac{\langle x,y\rangle}{\langle y,y\rangle}y$, then $\langle x,y \rangle = ||x||||y||$

Now attempt derivation:

$||x+y||^2=\langle x+y,x+y\rangle=||x||^2+||y||^2+2\langle x,y\rangle$

Now assuming that $x=\lambda y$

$||x+y||^2=\langle x+y,x+y\rangle=||x||^2+||y||^2+ 2||x|||y||$

Now I do not know how to proceed.

Question:

Am I on the right track? If so, then how should I proceed?

Thanks in advance!

Answer 1

Divide by $||x||+||y||$ and we want to have $$||z+w||=1$$ Where $||z|| + || w||=1$.

So, using your derivation

\begin{align*} ||z+w||^2&=||z||^2+||w||^2 +2<z,w>\\ &\leq ||z||^2+||w||^2 +2||z||||w||\\ &=(||z||+||w||)^2 =1 \end{align*} Therefore, $<z,w>=||z|| ||w||$.

Since the Cauchy-Schwartz equality holds if and only $z,w$ are linearly dependent, we get that $x=\lambda y$.

So, $$||x+y|| =|(1+\lambda)||x||,\quad ||x||+||y||=(1+|\lambda|)||x||$$ If $||x||\neq 0$, we obtain $\lambda = |\lambda |$, hence $\lambda >0$.

To sum up, the inequality holds if and only if $x=\lambda y$ where $\lambda \geq 0$.