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A convex function is defined as one that satisfies the following condition for $p_1 + p_2 = 1$.

$$f(p_1x_1 + p_2x_2) \leq p_1f(x_1) + p_2f(x_2),$$

Does this imply that for all $\lambda \leq 1$

$$f(\lambda x) \leq \lambda f(x)$$

I can imagine this is true but I can only show it for the case where $f(0) = 0$. Moreover, is the condition $\lambda\leq 1$ even necessary in the above statement? Essentially, I am looking for the most general statement one can make if $f$ is convex regarding $f(\lambda x)$.

EDIT: As pointed out by Michael, I should also add $0\leq \lambda$. My statement now is for $f$ convex such that $f(0) = 0$, and $0\leq \lambda \leq 1$ we have $f(\lambda x) \leq \lambda f(x)$. Is this the most general statement one can make?

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  • $\begingroup$ Consider $f(x)=e^x$ at $x=0$ for any $\lambda \in (0,1)$. $\endgroup$ – TM Gallagher Apr 14 at 17:28
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It's wrong. Try $f(x)=x^2$ and $\lambda=-1.$

If $0\leq \lambda\leq1$, so take $f(x)=e^x$.

If also $f(0)=0$ it's true already.

$$\lambda f(x)+(1-\lambda)f(0)\geq f(\lambda x+(1-\lambda)0),$$ which gives, which you wish.

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  • $\begingroup$ Ah excellent point! If I restrict it to $0 \leq \lambda \leq 1$, is my statement then true? $\endgroup$ – user1936752 Apr 14 at 17:30
  • $\begingroup$ @user1936752 If so, it's wrong for $f(x)=e^x$. We'll obtain $\lambda e^{(1-\lambda)x}\geq1,$ which is wrong. $\endgroup$ – Michael Rozenberg Apr 14 at 17:40
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    $\begingroup$ @user1936752 I added something. See please. $\endgroup$ – Michael Rozenberg Apr 14 at 17:52
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$f(x)=x^2+1$ (or even just $f(x)=1$) is convex but does not have your property.

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