The problem:
Assume $f:R^n→R$ is a convex function. Show that the function $g(t)=f(x_0+td), t∈R$ is convex for arbitrary $d, x_0∈R^n$. Explain the statement graphically when $n-2$.
My attempt at the solution:
To show that $g(x)$ is convex I have to use the convexity inequality.
$g(αt_1+(1-α)t_2)≤αg(t_2)+(1-α)g(t_2)$
Using the formula for $g(t)$ that was given we get
$f(α(x_0+t_1d)+(1-α)(x_0+t_2d))≤αf(x_0+t_1d)+(1-α)f(x_0+t_2d)$
After changing the left hand side I get something like this
$f(αt_1d+x_0+t_2d-αt_2d)≤αf(x_0+t_1d)+(1-α)f(x_0+t_2d)$
I'm stuck here. How can I make this look like a convexity formula for $f$?
Let $t_1,t_2\in\mathbb{R}$. Then \begin{align} g(\alpha t_1 + (1-\alpha) t_2) &= f(x_0+[\alpha t_1 + (1-\alpha) t_2]d) \\ &=f(\alpha(x_0+t_1 d)+(1-\alpha)(x_0+t_2 d))\\ &\leq \alpha f(x_0+t_1d)+(1-\alpha)f(x_0+t_2 d)\\ &= \alpha g(t_1)+(1-\alpha)g(t_2). \end{align} Then, $g$ satisfies the required inequality.
It seems as though you are supposing the equality is true, and then working backwards. This can be useful for "reverse engineering" a proof, but it's not really a proof itself.
What you need to do is take one side of the inequality, like $g(\alpha t_1 + (1-\alpha) t_2)$, and to make individual steps you know to be true to establish inequalities and eventually obtain the other side ($\alpha g(t_1)+(1-\alpha)g(t_2)$). You don't want to just suppose the inequality holds outright.
The proof is actually the same as for the more general case which I linked.