Inequality leading to Holder's: $t^\theta \leq \theta t + 1 - \theta$ with $0<\theta<1$ and $t\geq 0$

Question

In the process of proving Holder's Inequality for $l^p$ spaces, as per my instructions it begins by first asking us to prove the following inequality as a first step:

If $0<\theta<1$ and $t\geq 0$ then $$t^\theta \leq \theta t + (1-\theta)$$ and equality only holds if $t=1$.

The inequality is trivially true when $t=0$ and when $t=1$ but for other values of $t$ it is unclear to me how to approach this. Note also: $\theta t + (1-\theta) = 1 + (\theta-1)t$

My thought process:

• Attempt to bring the exponent down via logarithms, then $\theta \ln t \leq^? \ln(\theta t + (1-\theta))$, but attempting to expand the RHS with powerseries for $\ln$ seems unwieldy and equally difficult.
• Attempt to expand the LHS as $t^\theta = (1+(t-1))^\theta = \sum_k \binom{\theta}{k}1^{\theta-k}(t-1)^k$ where the coefficients $\binom{\theta}{k}$ are the generalized binomial coefficients $\binom{\theta}{k}:=\frac{\theta(\theta-1)(\theta-2)\cdots(\theta-k+1)}{k!}$. This also seems rather unwieldy and difficult to work with.
• Since we know it works for $t=0$ and for $t=1$, try to find a range such that $f(t,\theta) = \theta t + (1-\theta) - t^\theta>0$ for $t=0$ and for fixed theta that $\frac{\partial f}{\partial t} \geq 0$. We have $\frac{\partial f}{\partial t} = \theta -\theta t^{\theta-1}$ which for $t\geq 1$ since $-1<\theta-1<0$ will be $\theta - \theta\cdot \frac{1}{c}\geq 0$ where $c=t^{1-\theta}\geq t^0=1$ (with strict inequality for $t>1$)

This third approach seems to work for the case where $t\geq 1$ showing that difference between the RHS and the LHS is increasing and nonnegative for every value of $\theta$, but when $0<t<1$, you have $t^\theta>t$ so it appears that the derivative is in fact decreasing on the interval so it seems to require a different approach.

Where should I go from here to prove the case with $0\leq t\leq 1$?

---

aside: The form of Holder's Inequality that I am approaching is: if $x\in l^p$ and $y\in l^{p'}$ where $1\leq p\leq \infty$ and $\frac{1}{p}+\frac{1}{p'}=1$ then for $xy=(x_ky_k)$ you have $\|xy\|_1\leq \|x\|_p\|y\|_{p'}$. Where $1<p<\infty$ this is equivalent to writing: $\sum_k|x_ky_k|\leq\left(\sum_k|x_k|^p\right)^{\frac{1}{p}}\left(\sum_k|y_k|^{p'}\right)^{\frac{1}{p'}}$

Answer 1

For $t > 0$, write

$$t^\theta = t^\theta\cdot 1^{1-\theta}.$$

Taking the logarithm of both sides, what you need to show becomes

$$\theta\log t + (1-\theta) \log 1 \leqslant \log (\theta\cdot t + (1-\theta)\cdot 1),$$

which follows from the concavity of $\log$.

Answer 2

If we take $g:\theta\to t^{\theta}$ we are just stating that the graphics of $g$ on the interval $(0,1)$ lies below the line through $(0,1)$ and $(1,g(1))$. However, that is trivial, since $g(\theta)$ is a convex function due to: $$ g''(\theta) = t^{\theta}\log^2 t \geq 0.$$

Answer 3

Note that this is a special case of the general theorem of means.

If $a, b$ are non-negative and $0 < x, y < 1$ with $x + y = 1$ then $a^{x}b^{y} \leq ax + by$. Equality occurs if $a = b$.

Here in the current question $a = t, b = 1, x = \theta, y = 1 - \theta$. An easy proof of the general theorem mentioned above is based on Mean Value Theorem.

Clearly if $a = b$ then there is equality. So let's suppose $a < b$ (since the inequality does not change by interchanging $a$ and $b$ there is no need to consider $b > a$). Consider the function $f(t) = t^{1 - x}$ so that $f'(t) = (1 - x)t^{-x}$. By mean value theorem we get $$b^{1 - x} - a^{1 - x} = f(b) - f(a) = (b - a)f'(c) = (b - a)(1 - x)c^{-x}$$ where $a < c < b$. And since $-x < 0$ we have $c^{-x} < a^{-x}$. It follows that we have $$b^{1 - x} - a^{1 - x} < (b - a)(1 - x)a^{-x}$$ Multiplying by $a^{x} > 0$ and noting that $1 - x = y$ we get $$a^{x}b^{y} - a < (b - a)y$$ or $$a^{x}b^{y} < a(1 - y) + by = ax + by$$ This can be further generalized to the following theorem:

If $a_{1}, a_{2}, \ldots, a_{n}$ are non-negative and $x_{1}, x_{2}, \ldots, x_{n}$ are positive with $\sum\limits_{i = 1}^{n} x_{i} = 1$ then we have $$\prod_{i = 1}^{n}a_{i}^{x_{i}}\leq \sum_{i = 1}^{n}a_{i}x_{i}$$ and equality occurs only when all the $a_{i}$ are equal.