Suppose $x$ and $y$ are two vectors in $l_p$ (where $1\leq p<\infty$) such that $||x||=||y||=1$. Can we find a complex scalar $\alpha$, with $|\alpha|<1$ such that $||x+\alpha y||>1$. I don't know the answer to this even in finite dimensional $l_p^n$. For example, in $l_1^n$ thsi can be rephrased as:
Given $\displaystyle \sum_{i=1}^n|a_i|=\sum_{i=1}^n|b_i|=1$, can we find $|\alpha|<1$ such that $\displaystyle \sum_{i=1}^n|a_i+\alpha b_i|>1$?
On
If $1<p<\infty$ we've seen it's just strict convexity. The result is false for $p=\infty$. For $p=1$ there's a sort of complex version of strict convexity; although the unit sphere of $L^1$ does contain line segments it does not contain any (flat) complex disks:
Lemma. If $z,w\in\Bbb C$ and $w\ne0$ then $\frac1{2\pi}\int_0^{2\pi}|z+e^{it}w|\,dt>|z|$.
Proof: This is clear for $z=0$. If $z\ne0$ then wlog $z=1$ and then wlog $w=r>0$. The integral is equal to $\frac1{2\pi}\int_0^{2\pi}\phi(t)\,dt$, where $$\phi(t)=\frac 12(|1+e^{it}r|+|1-e^{it}r|).$$The triangle inequality shows that $\phi(t)\ge1$, and it's clear that $\phi(\pi/2)>1$, hence the integral is greater than $1$ since $\phi$ is continuous.
Cor. If $f,g\in L^1(\mu)$ and $g\ne0$ then $\frac1{2\pi}\int_0^{2\pi}||f+e^{it}g||_1\,dt>||f||_1$.
(So there exists $\alpha$ with $|\alpha|=1$ and $||f+\alpha g||_1>||f||_1$, and hence you can also get $|\alpha|<1$.)
Suppose $X$ is an arbitrary normed linear space, and $\|x\| = 1$. Then, at least one of $\|x + y\|$ and $\|x - y\|$ must be greater than or equal to $1$, as $$\|x + y\| + \|x - y\| \ge \|2x\| = 2.$$ To get strict inequality, assume further that $y \neq 0$ and $X$ is a strictly convex space, meaning that the unit sphere of $X$ contains no (non-degenerate) line segments (e.g. $l_p$ spaces for $1 < p < \infty$). Equivalently, if $v, w$ have norm $1$, then $\frac{v + w}{2}$ has norm strictly less than $1$.
If $\|x + y\|, \|x - y\| \le 1$, then $$2 \ge \|x + y\| + \|x - y\| \ge \|2x\| = 2,$$ which means that $\|x + y\| = \|x - y\| = 1$. Note that $$x = \frac{(x + y) + (x - y)}{2}.$$ This contradicts strict convexity. Thus, in a strictly convex space, $\|x + y\| > 1$ or $\|x - y\| > 1$.
EDIT: This property fails if and only if $X$ has a subspace that is isometrically isomorphic to $\ell^2_\infty$. Fix $\|x\| = \|y\| = 1$, and suppose $\|x + \alpha y\| \le 1$ for all $|\alpha| < 1$.
From the above methods, we can conclude that $\|x + \alpha y\| = 1$ for all $|\alpha| < 1$, as if $\|x + \alpha y\| < 1$, then $\|x - \alpha y\| > 1$. By continuity, this implies $\|x + \alpha y\| = 1$ for $|\alpha| = 1$ too.
Now, if $|\alpha| = 1$, then $|\alpha^{-1}| = 1$, so $\|y + \alpha x\| = |\alpha| \|x + \alpha^{-1}y\| = 1$. Note that $y$ is the midpoint of $y + \alpha x$ and $y - \alpha x$, all of which have norm $1$, so every point on the line segment between $y + \alpha x$ and $y - \alpha x$ have norm $1$. That is, if $|\alpha| \le 1$, we have $\|y + \alpha x\| = \|x + \alpha y\| = 1$.
Now, it's easy to see that $\|\alpha x + \beta y\| = \max\{|\alpha|, |\beta|\}$, by simply dividing by $\alpha$ or $\beta$; whichever has largest modulus. Our linear isometry is defined by $x \mapsto (1, 0)$ and $y \mapsto (0, 1)$ (note that $x \neq y$, as $x = y$ would imply that $\|x + 2y\| = 3$, not $2$).
So, all we need to do is show that $\ell_1$ does not contain such a subspace (this will also prove the property for $\ell_1^n$, as it can be identified with a subspace of $\ell_1$). Certainly in the real case, such a subspace does exist: $\ell_1^2$ (map $(1/2, 1/2)$ to $(1, 0)$ and $(1/2, -1/2)$ to $(0, 1)$), so the property fails. For the complex case, I don't know.