Show that $(f_1of_2)(x)=\inf \{f_1(x)+f_2(x):x=x_1+x_2\}$ is convex function

Asked Jun 08 '21 at 02:24

Active Jun 08 '21 at 02:34

Viewed 45 times

Show that if $f_1(x_1), f_2(x_2)$ are convex functions then ,

$(f_1of_2)(x)=\inf \{f_1(x)+f_2(x):x=x_1+x_2\}$ is convex function

The definition of a convex function f is that

$$f((1-t)x+ty)\le (1-t)f(x)+tf(y)\tag{1}$$

for all vectors $x,y\in \mathbb{R}^n,t \in [0,1]$

so,

$(f_1of_2)((1-t)x+ty)=(f_1of_2)\{((1-t)x_1+ty_1)+((1-t)x_2+ty_2)\}$

$=\inf\{((1-t)x_1+ty_1)+((1-t)x_2+ty_2):x=x_1+x_2,y=y_1+y_2\}$

I am just stuck weather I can use Infimum properties or not someone can help ?

edited Jun 08 '21 at 02:34

Arctic Char

asked Jun 08 '21 at 02:24

Inverse Problem

Are you assuming that the functions are bounded below? Otherwise the infimum may be $-\infty$. – Martin R Jun 08 '21 at 18:45
1

Do you mean $(f_1of_2)(x)=\inf {f_1(x_1)+f_2(x_2):x=x_1+x_2}$ ? – Martin R Jun 08 '21 at 19:49
@MartinR..NO SIR – Inverse Problem Jun 09 '21 at 07:27
NO what? It looks like the same question and has an answer. – Perhaps you can also respond to my first two comments. – Martin R Jun 09 '21 at 07:28
Btw, this operation is called “infimal convolution.” – Martin R Jun 09 '21 at 07:44
@MartinR.... I don't think so both are the same answers sir – Inverse Problem Jun 09 '21 at 07:55
Sorry, I do not understand what you are saying. Do you think that https://math.stackexchange.com/q/1244212 is a different question? Why? – Martin R Jun 09 '21 at 07:58
@MartinR sorry sir I'm so stressed with other things so can't pay attention to your comments yes sir both are the same..thank you so much for you patience – Inverse Problem Jun 09 '21 at 08:48

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