Show that $f(x)=x^4$ is convex

Question

for $x\in (0,\infty)$ show $f(x)=x^4$ is convex. I know it is convex since $f''(x)>0$ . How can we show by using definition? do we have to use Let L be linear space. $t\in[0,1],y\in L,f(xt+y(1-t))=(xt)^4+4(xt)^3((1-t)y)^1+6(xt)^2((1-t)y)^2+4(xt)(((1-t)y)^3+((1-t)y)^4$

edit: $(xt)^4+4(xt)^3((1-t)y)^1+6(xt)^2((1-t)y)^2+4(xt)(((1-t)y)^3+((1-t)y)^4\le tf(x)+4tf(x)+10tf(x)(1-t)f(y)+(1-t)f(y)$

Hint: $x^4 = (x^2)^2$ and $f(x) = x^2$ is convex and increasing on $\mathbb{R}_+$. — Random Jack, Aug 05 '14 at 20:41
http://math.stackexchange.com/questions/580856/proof-of-convexity-of-fx-x2 Then use the Hint of Random Jack :) — Bman72, Aug 05 '14 at 20:47
@ThomasAndrews I know. thanks. @ RandomJack goint to try soon — lyme, Aug 05 '14 at 20:53
@RandomJack I understand why $x^2$ is convex. is it true $((xt+(1-t)y)^2)^2\le (x^2t+(1-t)y^2)^2\le x^4t+(1-t)y^4+2(1-t)ty^2x^2$ how to continue :) — lyme, Aug 05 '14 at 21:51

score 4 · Answer 1 · answered Aug 05 '14 at 21:00

Note that by definition a function is convex in a domain when for any two points in the domain, the value of the linear interpolation between those two points for a third point in between them is greater than the value of the function at that third point. The idea is enter image description here Now all you have to check is that for any ${x_{\rm{i}}} > 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabdIha4naaBaaaleaacqqGPbqA % aeqaaOGaeyOpa4JaeGimaadaaa!43FD! $ and ${x_{\rm{f}}} > {x_{\rm{i}}} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabdIha4naaBaaaleaacqqGMbGz % aeqaaOGaeyOpa4JaemiEaG3aaSbaaSqaaiabbMgaPbqabaaaaa!4607! $ and $\xi = (1 - t){x_{\rm{i}}} + t{x_{\rm{f}}} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabe67a4jabg2da9iabcIcaOiab % igdaXiabgkHiTiabdsha0jabcMcaPiabdIha4naaBaaaleaacqqGPb % qAaeqaaOGaey4kaSIaemiDaqNaemiEaG3aaSbaaSqaaiabbAgaMbqa % baaaaa!4F1B! $ (such that $0 < t < 1 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabicdaWiabgYda8iabdsha0jab % gYda8iabigdaXaaa!4456! $), $$f({x_{\rm{i}}}) + \frac{{f({x_{\rm{f}}}) - f({x_{\rm{i}}})}}{{{x_{\rm{f}}} - {x_{\rm{i}}}}}(\xi - {x_{\rm{i}}}) > f(\xi ) % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabdAgaMjabcIcaOiabdIha4naa % BaaaleaacqqGPbqAaeqaaOGaeiykaKIaey4kaSYaaSaaaeaacqWGMb % GzcqGGOaakcqWG4baEdaWgaaWcbaGaeeOzaygabeaakiabcMcaPiab % gkHiTiabdAgaMjabcIcaOiabdIha4naaBaaaleaacqqGPbqAaeqaaO % GaeiykaKcabaGaemiEaG3aaSbaaSqaaiabbAgaMbqabaGccqGHsisl % cqWG4baEdaWgaaWcbaGaeeyAaKgabeaaaaGccqGGOaakcqaH+oaEcq % GHsislcqWG4baEdaWgaaWcbaGaeeyAaKgabeaakiabcMcaPiabg6da % +iabdAgaMjabcIcaOiabe67a4jabcMcaPaaa!6738! $$. Replacing the function that you have asked yields proving that $${x_{\rm{f}}}^4 - ((1 - t){x_{\rm{i}}}^4 + t{x_{\rm{f}}}^4) + t({x_{\rm{f}}}^4 - {x_{\rm{i}}}^4) > 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfKttLearuqr1ngBPrgarmqr1ngBPrgitL % xBI9gBamXvP5wqSXMqHnxAJn0BKvguHDwzZbqegm0B1jxALjhiov2D % aeHbuLwBLnhiov2DGi1BTfMBaebbfv3ySLgzGueE0jxyaibaieYlf9 % irVeeu0dXdh9vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqaq-J % frVkFHe9pgea0dXdar-Jb9hs0dXdbPYxe9vr0-vr0-vqpWqaaeaabi % GaciaacaqabeaadaabauaaaOqaaiabdIha4naaBaaaleaacqqGMbGz % aeqaaOWaaWbaaSqabeaacqaI0aanaaGccqGHsislcqGGOaakcqGGOa % akcqaIXaqmcqGHsislcqWG0baDcqGGPaqkcqWG4baEdaWgaaWcbaGa % eeyAaKgabeaakmaaCaaaleqabaGaeGinaqdaaOGaey4kaSIaemiDaq % NaemiEaG3aaSbaaSqaaiabbAgaMbqabaGcdaahaaWcbeqaaiabisda % 0aaakiabcMcaPiabgUcaRiabdsha0jabcIcaOiabdIha4naaBaaale % aacqqGMbGzaeqaaOWaaWbaaSqabeaacqaI0aanaaGccqGHsislcqWG % 4baEdaWgaaWcbaGaeeyAaKgabeaakmaaCaaaleqabaGaeGinaqdaaO % GaeiykaKIaeyOpa4JaeGimaadaaa!64D0! $$. Note that the above inequality is trivially true when the domain is restricted to the positive real line, so the function is convex indeed.

Random Jack · Accepted Answer · 2014-08-05T22:23:15.550

It is easy to show that $f(x) = x^2$ is convex and increasing on $\mathbb{R}_+$.

Hence $\forall x, y \in \mathbb{R}_+, t \in [0, 1]$ we have:

$$(tx + (1-t)y)^4 = ((tx + (1-t)y)^2)^2 \stackrel{(1)}\leqslant (tx^2 + (1-t)y^2)^2 \stackrel{(2)}\leqslant \\ t(x^2)^2 + (1-t)(y^2)^2 = tx^4 + (1-t)y^4.$$

$(1)$: using that $x^2$ is convex and increasing.

$(2)$: again using that $x^2$ is convex.

Note also that in your question $L = \mathbb{R}_+$. This is not a linear space and it should not be. But it must be convex because we can speak about convexity of a function only on a convex subset of its domain.

Show that $f(x)=x^4$ is convex

2 Answers2