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Gordon
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Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\,\beta_t\, dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The caplet Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.

Applying a shift, the probability distribution has been changed, for example, lognormal distribution has been changed to a shifted lognormal distribution. However, the price will not change, while the implied volatility changes. In the current negative or small interest rate environment, people tend to quote an interest rate product by its price. Then given the price, an implied volatility is computed with a certain shift parameter; otherwise, it may not be feasible to find the implied volatility (e.g., the forward Libor rate $L(0; T, T+\Delta)$ may be negative). See also this paper.

Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\,\beta_t\, dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The caplet Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.

Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\,\beta_t\, dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The caplet Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.

Applying a shift, the probability distribution has been changed, for example, lognormal distribution has been changed to a shifted lognormal distribution. However, the price will not change, while the implied volatility changes. In the current negative or small interest rate environment, people tend to quote an interest rate product by its price. Then given the price, an implied volatility is computed with a certain shift parameter; otherwise, it may not be feasible to find the implied volatility (e.g., the forward Libor rate $L(0; T, T+\Delta)$ may be negative). See also this paper.

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Gordon
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Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\beta_t dW_t, \quad 0 < t \le T, \end{align*}\begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\,\beta_t\, dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The optioncaplet Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.

Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\beta_t dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The option Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.

Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\,\beta_t\, dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The caplet Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.

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Gordon
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Let \begin{align*} L(t; T, T + \Delta) = \frac{1}{\Delta} \left[ \frac{P(t,T)}{P(t, T+\Delta)} - 1 \right] \end{align*} be the forward Libor rate at time $t$ for the period $[T, T+\Delta]$. Consider a caplet with payoff at $T+\Delta$ of the form \begin{align} \Delta\max\big(L(T; T, T + \Delta) -K, \, 0 \big) &= \Delta\max\big((L(T; T, T + \Delta)-\alpha) -(K-\alpha), \, 0 \big) \tag{1} \end{align} We assume that \begin{align*} L(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta) + \alpha, \quad 0 < t \le T, \end{align*} where the process $\hat{L} = \{ \hat{L}(t; T, T + \Delta) \, | \, 0 < t \le T\}$ satisfies an SDE of the form \begin{align*} d \hat{L}(t; T, T + \Delta) = \hat{L}(t; T, T + \Delta)\beta_t dW_t, \quad 0 < t \le T, \end{align*} under the $T+\Delta$-forward probability measure, where $\beta$ is a deterministic function, and $\{W_t \mid t > 0\}$ is a standard Brownian motion. The option Payoff (1) then has value \begin{align} C(\alpha, \sigma) = P(0, T+\Delta) \Delta \Big[\big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\Big],\tag{2} \end{align} where \begin{align*} d_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \pm \frac{1}{2}\sigma^2(0, T) T}{\sigma(0, T) \sqrt{T}}, \end{align*} and \begin{align*} \sigma(0, T) = \sqrt{\frac{1}{T}\int_0^T \beta^2_t dt}. \end{align*} The implied volatility $\hat{\sigma}$ is a quantity such that \begin{align*} L(0; T, T + \Delta) \Phi(\hat{d}_1) - K \Phi(\hat{d}_2) = \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2),\tag{3} \end{align*} where \begin{align*} \hat{d}_{1, 2} = \frac{\ln \frac{L(0; T, T + \Delta)}{K} \pm \frac{1}{2}\hat{\sigma}^2 T}{\hat{\sigma} \sqrt{T}}. \end{align*} Given $T$ and $K$, the implied volatility $\hat{\sigma}$ is a function of $\alpha$.

Let $f(\alpha)$ be the right hand side of (3), that is, \begin{align*} f(\alpha) &= \big(L(0; T, T + \Delta)-\alpha\big) \Phi(d_1) - (K-\alpha) \Phi(d_2)\\ &=(K-\alpha)\bigg[\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) - \Phi(d_2)\bigg]. \end{align*} Then, \begin{align*} \frac{df(\alpha)}{d\alpha} &= -\frac{L(0; T, T + \Delta)-\alpha}{K-\alpha} \Phi(d_1) + \Phi(d_2) + \Phi(d_1) \frac{L(0; T, T + \Delta)-K}{K-\alpha}\\ &=\Phi(d_2) - \Phi(d_1) < 0. \end{align*} The derivative with respect to $\alpha$ of the left hand side of (3) is given by \begin{align*} vega \times \frac{d \hat{\sigma}}{d \alpha}. \end{align*} That is, \begin{align*} \frac{d \hat{\sigma}}{d \alpha} < 0. \end{align*} In other words, increasing $\alpha$ shifts the implied volatility curve $\hat{\sigma}(K)$ down, while decreasing $\alpha$ shifts the curve up.